Class 12 - Maths - Three Dimensional Geometry

                                                                     Exercise 11.1

Question 1:

If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.

Answer:

Let direction cosines of the line be l, m, and n.

l = cos 90° = 0

m = cos 135° = -1/√2

n = cos 45° = 1/√2

Therefore, the direction cosines of the line are 0, 1/√2 and 1/√2

Question 2:

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer:

Let the direction cosines of the line make an angle α with each of the coordinate axes.

l = cos α, m = cos α, n = cos α

Now, l2 + m2 + n2 = 1

=> cos2 α + cos2 α + cos2 α = 1

=> 3 cos2 α = 1

=> cos2 α = 1/3

=> cos α = ±1/√3

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes are

±1/√3, ±1/√3 and ±1/√3

Question 3:

If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

Answer:

If a line has direction ratios of −18, 12, and −4, then its direction cosines are

    -18/√{(-18)2 + (12)2 + (-4)2}, 12/√{(-18)2 + (12)2 + (-4)2}, -4/√{(-18)2 + (12)2 + (-4)2}

= -18/√(324 + 144 + 16), 12/√(324 + 144 + 16), -4/√(324 + 144 + 16)

= -18/√(484), 12/√(484), -4/√(484)

= -18/22, 12/22, -4/22

Thus, the direction cosines are: -18/22, 12/22 and -4/22

Question 4:

Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Answer:

The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).

It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2),

are given by, x2 − x1, y2 − y1, and z2 − z1.

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.

It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional.

Therefore, AB is parallel to BC.

Since point B is common to both AB and BC, points A, B and C are collinear.

Question 5:

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (−1, 1, 2) and (− 5, − 5, − 2).

Answer:

The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

 

                                                             

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

Then, √{(-4)2 + (-4)2 + (6)2} = √(16 + 16 + 36) = √(68) = 2√(17)

Therefore, the direction cosines of AB are

   -4/√{(-4)2 + (-4)2 + (6)2}, -4/√{(-4)2 + (-4)2 + (6)2}, 6/√{(-4)2 + (-4)2 + (6)2}

= -4/2√(17), -4/2√(17), 6/2√(17)

= -2/√(17), -2/√(17), 3/√(17)

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.

Therefore, the direction cosines of BC are

   -4/√{(-4)2 + (-4)2 + (6)2}, -6/√{(-4)2 + (-4)2 + (6)2}, -4/√{(-4)2 + (-4)2 + (6)2}

= -4/2√(17), --6/2√(17), -4/2√(17)

= -2/√(17), -3/√(17), -2/√(17)

The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2.

Therefore, the direction cosines of AC are

    -8/√{(-8)2 + (10)2 + (2)2}, -10/√{(-8)2 + (10)2 + (2)2}, 2/√{(-8)2 + (10)2 + (2)2}

= -8/√(64 + 100 + 4), -10/√(64 + 100 + 4), 2/√(64 + 100 + 4)

= -8/√(168), -10/√(169), 2/√(168)

= -8/2√(42), -10/2√(42), 2/2√(42)

= -4/√(42), -5/√(42), 1/√(42)

                                                                   Exercise 11.2

Question 1:

Show that the three lines with direction cosines

12/13, -3/13, -4/13, 4/13, 12/13, 3/13, 3/13, -4/13, 12/13

are mutually perpendicular.

Answer:

Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each other, if

l1l2 + m1m2 + n1n2 = 0

(i) For the lines with direction cosines 12/13, -3/13, -4/13 and 4/13, 12/13, 3/13

l1l2 + m1m2 + n1n2 = (12/13) * (4/13) + (-3/13) * (12/13) + (-4/13) * (3/13)

                                = 48/169 – 36/169 – 12/169

                                = 48/169 – 48/169

                                = 0               

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines 4/13, 12/13, 3/13 and 3/13, -4/13, 12/13

l1l2 + m1m2 + n1n2 = (4/13) * (3/13) + (12/13) * (-4/13) + (3/13) * (12/13)

                                = 12/169 - 48/169 + 36/169

                                = 48/169 – 48/169

                                = 0               

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines 3/13, -4/13, 12/13 and 12/13, -3/13, -4/13

l1l2 + m1m2 + n1n2 = (3/13) * (12/13) + (-4/13) * (-3/13) + (12/13) * (-4/13)

                                = 36/169 + 12/169 – 48/169

                                = 48/169 – 48/169

                                = 0               

Thus, all the lines are mutually perpendicular.

Question 2:

Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Answer:

Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the

points, (0, 3, 2) and (3, 5, 6).

The direction ratios, a1, b1, c1 of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5 and −4.

The direction ratios, a2, b2, c2 of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.

AB and CD will be perpendicular to each other, if a1a2 + b1b2+ c1c2 = 0

Now, a1a2 + b1b2+ c1c2 = 2 * 3 + 5 * 2 + (− 4) * 4

                                        = 6 + 10 − 16

                                        = 0

Therefore, AB and CD are perpendicular to each other.

Question 3:

Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

Answer:

Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the

points, (−1, −2, 1) and (1, 2, 5).

The directions ratios, a1, b1, c1 of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.

The direction ratios, a2, b2, c2 of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4.

AB will be parallel to CD, if a1/a2 = b1/b2 = c1/c2

a1/a2 = -2/2 = -1

b1/b2 = -4/4 = -1

c1/c2 = -4/4 = -1

So, a1/a2 = b1/b2 = c1/c2

Thus, AB is parallel to CD.

Question 4:

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3i + 2j – 2k.

Answer:

It is given that the line passes through the point A (1, 2, 3).

Therefore, the position vector through A is

a = i + 2j + 3k

b = 3i + 2j – 2k

It is known that the line which passes through point A and parallel to b is given by

      r = a + λb, where λ is a constant.   

=> r = (i + 2j + 3k) + λ (3i + 2j – 2k)

This is the required equation of the line.

Question 5:

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i – j + 4k and is in the direction i + 2j – k.

Answer:

It is given that the line passes through the point with position vector

a = 2i – j + 4k   ………….1

b = i + 2j – k   ………….2

It is known that a line through a point with position vector a and parallel to b is given by

      r = a + λb, where λ is a constant.

=> r = (2i - j + 4k) + λ(i + 2j – k)

This is the required equation of the line in vector form.

r = xi – yj + zk

=> xi – yj + zk = (2i - j + 4k) + λ(i + 2j – k)

=> xi – yj + zk = (λ + 2)i + (2λ - 1)j + (-λ + 4)k

Eliminating λ, we obtain the Cartesian form equation as

(x - 2)/1 = (y + 1)/2 = (z - 4)/(-1)

This is the required equation of the given line in Cartesian form.

Question 6:

Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by (x + 3)/3 = (y - 4)/5 = (z + 8)/6

Answer:

It is given that the line passes through the point (−2, 4, −5) and is parallel to

(x + 3)/3 = (y - 4)/5 = (z + 8)/6

The direction ratios of the line (x + 3)/3 = (y - 4)/5 = (z + 8)/6 are 3, 5, and 6.

The required line is parallel to (x + 3)/3 = (y - 4)/5 = (z + 8)/6

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x1, y1, z1) and with direction ratios,

a, b, c, is given by (x – x1)/a = (y – y1)/b = (z – z1)/c

Therefore the equation of the required line is

      (x + 2)/3k = (y - 4)/5k = (z + 5)/6k

=> (x + 2)/3 = (y - 4)/5 = (z + 5)/6 = k

Question 7:

The Cartesian equation of a line is (x - 5)/3 = (y + 4)/7 = (z - 6)/2. Write its vector form.

Answer:

The Cartesian equation of the line is

(x - 5)/3 = (y + 4)/7 = (z - 6)/2    ………..1

The given line passes through the point (5, −4, 6). The position vector of this point is

a = 5i – 4j + 6k

Also, the direction ratios of the given line are 3, 7, and 2.

This means that the line is in the direction of vector, b = 3i + 7j + 2k

It is known that the line through position vector and in the direction of the vector is given by

the equation,

      r = a + λb, where λ is a constant.

=> r = (5i - 4j + 6k) + λ(3i + 7j + 2k)

This is the required equation of the given line in vector form.

Question 8:

Find the vector and the Cartesian equations of the lines that pass through the origin and          (5, −2, 3).

Answer:

The required line passes through the origin. Therefore, its position vector is given by,

a = 0     ………1

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation,

b = 5i – 2j + 3k

The equation of the line in vector form through a point with position vector and parallel to b is

      r = a + λb, where λ є R

=> r = 0 + λ(3i + 7j + 2k)

=> r = λ(3i + 7j + 2k)

It is known that the equation of the line through the point (x1, y1, z1) and with direction ratios,

a, b, c, is given by (x – x1)/a = (y – y1)/b = (z – z1)/c

Therefore, the equation of the required line in the Cartesian form is

      (x - 0)/5 = (y - 0)/(-2) = (z - 0)/3

=> x/5 = y/(-2) = z/3

Question 9:

Find the vector and the Cartesian equations of the line that passes through the points                 (3, −2, −5), (3, −2, 6).

Answer:

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

a = 3i – 2j – 5k

The direction ratios of PQ are given by,

(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11

The equation of the vector in the direction of PQ is

B = 0.i – 0.j + 11k = 11k

The equation of PQ in vector form is given by,

      r = a + λb, where λ є R

=> r = (3i - 2j - 5k) + 11λk

The equation of PQ in Cartesian form is

      (x – x1)/a = (y – y1)/b = (z – z1)/c

=> (x - 3)/0 = (y + 2)/0 = (z + 5)/11

Question 10:

Find the angle between the following pairs of lines:

(i) r = 2i – 5j + k + λ(3i – 2j + 6k) and r = 7i – 5k + μ(i + 2j + 2k)

(ii) r = 3i + j - 2k + λ(i – j - 2k) and r = 2i – j - 56k + μ(3i - 5j - 4k)

Answer:

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

cos Q = |(b1 . b2)/(|b1||b2|)|

The given lines are parallel to the vectors, b1 = 3i + 2j + 6k and b2 = i + 2j + 2k respectively.

Now, |b1| = √(32 + 22 + 62) = √(9 + 4 + 36) = √(49) = 7

And |b2| = √(12 + 22 + 22) = √(1 + 4 + 4) = √9 = 3

Now, b1 . b2 = (3i + 2j + 6k).(i + 2j + 2k)

                      = 3 * 1 + 2 * 2 + 6 * 2

                      = 3 + 4 + 12

                      = 19

So, cos Q = 19/(7 * 3)

=> cos Q = 19/21

=> Q = cos-1 (19/21)

(ii) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

cos Q = |(b1 . b2)/(|b1||b2|)|

The given lines are parallel to the vectors, b1 = i - j - 2k and b2 = 3i - 5j - 4k respectively.

Now, |b1| = √{12 + (-1)2 + (-2)2} = √(1 + 1 + 4) = √6

And |b2| = √{32 + (-5)2 + (-4)2} = √(9 + 25 + 16) = √(50) = 5√2

Now, b1 . b2 = (i - j - 2k).( 3i - 5j - 4k)

                      = 1 * 3 + 1 * 5 + 2 * 4

                      = 3 + 5 + 8

                      = 16

So, cos Q = 16/(√6 * 5√2)

=> cos Q = 16/(√2 * √3 * 5√2)

=> cos Q = 16/(10 * √3)

=> cos Q = 8/5√3

=> Q = cos-1 (8/5√3)

Question 11:

Find the angle between the following pairs of lines:

(i) (x - 2)/2 = (y - 1)/5 = (z + 3)/(-3) and (x + 2)/(-1) = (y - 4)/8 = (z - 5)/4

(ii) x/2 = y/2 = z/1 and (x - 5)/4 = (y - 2)/1 = (z - 3)/8

Answer:

(i) Let b1 and b2 be the vectors parallel to the pair of lines,

(x - 2)/2 = (y - 1)/5 = (z + 3)/(-3) and (x + 2)/(-1) = (y - 4)/8 = (z - 5)/4

So, b1 = 2i + 5j - 3k and b2 = -i + 8j + 4k

Now, |b1| = √{22 + 52 + (-3)2} = √(4 + 25 + 9) = √(38)

And |b2| = √{(-1)2 + 82 + 42} = √(1 + 64 + 16) = √(81) = 9

Now, b1 . b2 = (2i + 5j - 3k).(-i + 8j + 4k)

                      = -2 * 1 + 5 * 8 - 3 * 4

                      = -2 + 40 - 12

                      = 26

The angle, Q, between the given pair of lines is given by the relation,

      cos Q = |(b1 . b2)/(|b1||b2|)|

=> cos Q = 26/9√38

=> Q = cos-1 (26/9√38)

(ii) Let b1 and b2 be the vectors parallel to the pair of lines,

x/2 = y/2 = z/1 and (x - 5)/4 = (y - 2)/1 = (z - 3)/8

So, b1 = 2i + 2j + k and b2 = 4i + j + 8k

Now, |b1| = √{22 + 22 + (-1)2} = √(4 + 4 + 1) = √9 = 3

And |b2| = √{42 + 12 + 82} = √(16 + 1 + 64) = √(81) = 9

Now, b1 . b2 = (2i + 2j + k).(4i + j + 8k)

                      = 2 * 4 + 2 * 1 + 1 * 8

                      = 8 + 2 + 8

                      = 16

The angle, Q, between the given pair of lines is given by the relation,

      cos Q = |(b1 . b2)/(|b1||b2|)|

=> cos Q = 18/(3 * 9)

=> cos Q = 2/3

=> Q = cos-1 (2/3)

 

Question 12:

Find the values of p so the line (1 - x)/3 = (7y - 14)/2p = (z - 3)/2 and

(7 – 7x)/3p = (y - 5)/1 = (6 - z)/5 are at right angles.

Answer:

The given equations can be written in the standard form as

(x - 1)/(-3) = (y - 2)/(2p/7) = (z - 3)/2 and (x - 1)/(-3p/7) = (y - 5)/1 = (z - 6)/(-5)

The direction ratios of the lines are −3, 2p/7, 2 and -3p/7, 1, -5 respectively.

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if

      a1a2 + b1 b2 + c1c2 = 0

=> (-3) * (-3p/7) + (2p/7) * 1  +2 * (-5) = 0

=> 9p/7 + 2p/7 = 10

=> 11p/7 = 10

=> 11p = 70

=> p = 70/11

Thus, the value of p is 70/11

Question 13:

Show that the lines (x - 5)/7 = (y + 2)/(-5) = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.

Answer:

The equations of the given lines are (x - 5)/7 = (y + 2/(-5)) = z/1 and x/1 = y/2 = z/3

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other 

if 1a2 + b1 b2 + c1c2 = 0

Now, 7 * 1 + (−5) * 2 + 1 * 3 = 7 – 10 + 3 = 0

Therefore, the given lines are perpendicular to each other.

Question 14:

Find the shortest distance between the lines

r = i + 2j + k + λ(i – j + k) and r = 2i - j - k + μ(2i + j + 2k)

Answer:

The equations of the given lines are:

r = i + 2j + k + λ(i – j + k)

r = 2i - j - k + μ(2i + j + 2k)

It is known that the shortest distance between the lines r1 = a1 + λb1 and r2 = a2 + μb2 is given

by,

d = |{(b1 * b2).(a2 – a1)}/|b1 * b2||   ……………1

Comparing the given equations, we obtain

a1 = i + 2j + k

b1 = i - j + k

a2 = 2i - j - k

b2 = 2i + j + 2k

Now, a2 – a1 = (2i - j - k) – (i + 2j + k) = i - 3j - 2k

b1 * b2 =   i          j          k

                  1        -1        1    

                  2         1         2      

=> b1 * b2 = (-2 - 1)i – (2 - 2)j + (1 + 2)k = -3i + 3k

Now, |b1 * b2| = √{(-3)2 + 32} = √(9 + 9) = √(18) = 3√2

Substituting all the values in equation 1, we obtain

      d = |{(-3i + 3k).(i – 3j – 2k)}/3√2|

=> d = |(-3 * 1 – 3 * 2)/3√2|

=> d = |-9/3√2|

=> d = 3/√2

=> d = (3/√2) * (√2/√2)

=> d = 3√2/2

Therefore, the shortest distance between the two lines is 3√2/2 units.

Question 15:

Find the shortest distance between the lines

(x + 1)/7 = (y + 1)/(-6) = (z + 1)/1 and (x - 3)/1 = (y – 5)/(-2) = (z - 7)/1

Answer

The given lines are:

(x + 1)/7 = (y + 1)/(-6) = (z + 1)/1 and (x - 3)/1 = (y – 5)/(-2) = (z - 7)/1

It is known that the shortest distance between the two lines,

(x – x1)/a1 = (y – y1)/b1 = (z - z1)/c1 and (x – x2)/a2 = (y – y2)/b2 = (z - z2)/c2 is given by

           x2 – x1    y2 – y1     z2 – z1              

d =         a1            b1            c1                                         ……………………1

              a2            b2            c2    

        √{(b1c2 – b2c1)2 + (c1a2 – c2a1)2 + (a1b2 – a2b1)2}

Comparing the given equations, we obtain

x1 = -1, y1 = -1, z1 = -1

a1 = 7, b1 = -6, c1 = 1

x2 = 3, y2 = 5, z2 = 7

a2 = 1, b2 = -2, c2 = 1

Now,

           x2 – x1    y2 – y1     z2 – z1      =    4        6        8              

            a1            b1            c1                   7       -6        1                    

              a2            b2            c2                 1        -2       1    

= 4(-6 + 2) – 6(7 - 1) + 8(-14 + 6)

= -16 – 36 – 64

= -116

√{(b1c2 – b2c1)2 + (c1a2 – c2a1)2 + (a1b2 – a2b1)2} = √{(-6 + 2)2 + (1 + 7)2 + (-14 + 6)2}

                                                                                   = √(16 + 36 + 64)

                                                                                   = √116

                                                                                   = 2√29

Substituting all the values in equation 1, we obtain

      d = -116/2√29 = -58/√29

=> d = -(58/√29) * (√29/√29)

=> d = -58√29/29

=> d = -2√29

Since distance is always non-negative, the distance between the given lines is 2√29 units.

Question 16:

Find the shortest distance between the lines whose vector equations are

r = i + 2j + 3k + λ(i – 3j + 2k) and r = 4i + 5j + 6k + μ(2i + 3j + k)

Answer:

The equations of the given lines are:

r = i + 2j + 3k + λ(i – 3j + 2k) and r = 4i + 5j + 6k + μ(2i + 3j + k)

It is known that the shortest distance between the lines r1 = a1 + λb1 and r2 = a2 + μb2 is given

by,

d = |{(b1 * b2).(a2 – a1)}/|b1 * b2||   ……………1

Comparing the given equations, we obtain

a1 = i + 2j + 3k

b1 = i - 3j + 3k

a2 = 4i + 5j + 6k

b2 = 2i + 3j + k

Now, a2 – a1 = (4i + 5j + 6k) – (i + 2j + 3k) = 3i + 3j + 3k

b1 * b2 =   i          j          k

                  1        -3        2    

                  2         3         1      

=> b1 * b2 = (-3 - 6)i – (1 - 4)j + (3 + 6)k = -9i + 3j + 9k

Now, |b1 * b2| = √{(-9)2 + 32 + 92} = √(81 + 9 + 81) = √(171) = 3√19

Substituting all the values in equation 1, we obtain

      d = |{(-9i + 3j + 9k).( 3i + 3j + 3k)}/3√19|

=> d = |(-9 * 3 + 3 * 3 + 9 * 3)/3√19|

=> d = 9/3√19

=> d = 3/√19

Therefore, the shortest distance between the two lines is 3/√19 units.

 

 

Question 17:

Find the shortest distance between the lines whose vector equations are

r = (1 - t)i + (t - 2)j + (3 – 2t)k and r = (s + 1)i + (2s - 1)j - (2s + 1)k

Answer:

The equations of the given lines are:

      r = (1 - t)i + (t - 2)j + (3 – 2t)k

=> r = (i – 2j + 3k) + t(-I + j – 2k)    …………..1

      r = (s + 1)i + (2s - 1)j - (2s + 1)k

=> r = (i – j + k) + s(i + 2j – 2k)     …………….2

It is known that the shortest distance between the lines r1 = a1 + λb1 and r2 = a2 + μb2 is given

by,

d = |{(b1 * b2).(a2 – a1)}/|b1 * b2||   ……………3

Comparing the given equations, we obtain

a1 = i - 2j + 3k

b1 = -i + j - 2k

a2 = i - j - k

b2 = i + 2j - 2k

Now, a2 – a1 = (i - j - k) – (i - 2j + 3k) = j - 34k

b1 * b2 =   i          j          k

                 -1        1        -2    

                  1         2        -2      

=> b1 * b2 = (-2 + 4)i – (2 + 2)j + (-2 - 1)k = 2i - 4j - 3k

Now, |b1 * b2| = √{22 + (-4)2 + (-3)2} = √(4 + 16 + 9) = √(29)

Substituting all the values in equation 3, we obtain

      d = |{2i - 4j - 3k.(j - 4k)}/√29|

=> d = |(-4 + 12)/√29|

=> d = 8/√29

Therefore, the shortest distance between the two lines is 8/√29 units.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                                                                        Exercise 11.3

Question 1:

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a) z = 2                    (b) x + y + z = 1                       (c) 2x + 3y – z = 5                         (d) 5y + 8 = 0

Answer:

(a) The equation of the plane is z = 2 or 0.x + 0.y + z = 2    ………..1

The direction ratios of normal are 0, 0, and 1.

So, √(02+ 02 + 12) = 1

Dividing both sides of equation 1 by 1, we get

0.x + 0.y + 1.z = 2

This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the

plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is

2 units.

(b) Given, x + y + z = 1   …………1

The direction ratios of normal are 1, 1, and 1.

So, √(12+ 12 + 12) = √(1 + 1 + 1) = √3    

Dividing both sides of equation 1 by √3, we get

x/√3 + y/√3 + z/√3 = 1/√3    ………….2

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal

to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are 1/√3, 1/√3 and 1/√3 and the distance of

normal from the origin is 1/√3 units.

(c) Given, 2x + 3y − z = 5    ……………1

The direction ratios of normal are 2, 3, and −1.

So, √{22+ 32 + (-1)2} = √(4 + 9 + 1) = √14    

Dividing both sides of equation 1 by √14, we get

2x/√14 + 3y/√14 - z/√14 = 5/√14    ………….2

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal

to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 2/√14, 3/√14 and -1/√14 and

the distance of normal from the origin is 5/√14 units.

(d) Given, 5y + 8 = 0

0x − 5y + 0z = 8       ........... (1)

The direction ratios of normal are 0, −5, and 0.

So, √{02 + (-5)2 + 02} = 5    

Dividing both sides of equation 1 by 5, we get

-y = 8/5    ………….2

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal

to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of

normal from the origin is 8/5 units.

Question 2:

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i + 5j – 6k.

Answer:

The normal vector is n = 3i + 5j – 6k

Now, n^ = n/|n| = (3i + 5j – 6k)/√{32+ 52 + (-6)2} = (3i + 5j – 6k)/√70

It is known that the equation of the plane with position vector r is given by,

       r.n = d

=> r . {(3i + 5j – 6k)/√70} = 7

This is the vector equation of the required plane.

Question 3:

Find the Cartesian equation of the following planes:

(a) r.(i + j + k) = 2                 (b) r.(2i + 3j – 4k) = 1              (c) r.[(s – 2t)I + (3 - t)j + (2s + t)k] = 15

Answer:

(a) It is given that equation of the plane is

r.(i + j + k) = 2     ………..1

For any arbitrary point P (x, y, z) on the plane, position vector r is given by,

r = xi + yj + zk

Substituting the value of r in equation 1, we obtain

(xi + yj + zk)(i + j − k) = 2

=> x + y − z = 2

This is the Cartesian equation of the plane.

(b) Given, r. (2i + 3j – 4k) = 1     .......1

For any arbitrary point P (x, y, z) on the plane, position vector r is given by,

r = xi + yj + zk

Substituting the value of r in equation 1, we obtain

(xi + yj + zk)(2i + 3j – 4k) = 1

=> 2x + 3y − 4z = 2

This is the Cartesian equation of the plane.

(c) r.[(s − 2t)I + (3 − t)j + (2s + t)k] = 15     .......1

For any arbitrary point P (x, y, z) on the plane, position vector r is given by,

r = xi + yj + zk

Substituting the value of r in equation 1, we obtain

      (xi + yj + zk)[(s − 2t)i + (3 − t)j + (2s + t)k] = 15

=> (s − 2t)x + (3 − t)y + (2s + t)z = 15

This is the Cartesian equation of the plane.

Question 4:

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) 2x + 3y + 4z – 12 = 0         (b) 3y + 4x – 6 = 0           (c) x + y + z = 1         (d) 5y + 8 = 0

Answer:

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

Given, 2x + 3y + 4z − 12 = 0

=> 2x + 3y + 4z = 12        ..........1

The direction ratios of normal are 2, 3, and 4.

So, √(22 + 32 + 42} = √(4 + 9 + 16) = √29

Dividing both sides of equation 1 by √29, we get

2x/√29 + 3y/√29 + 4z/√29 = 12/√29

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of

normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd). Therefore,

the coordinates of the foot of the perpendicular are

{(2/√29) * (12/√29), (3/√29) * (12/√29), (4/√29) * (12/√29)} i.e. (24/29, 36/29, 48/29)

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

Given, 3y + 4z – 6 = 0

=> 0x + 3y + 4z = 6     ……………..1

The direction ratios of the normal are 0, 3, and 4.

So, √(02 + 32 + 42) = √(0 + 9 + 16) = √25 = 5

Dividing both sides of equation 1 by 5, we obtain

0x + 3y/5 + 4z/5 = 6/5

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal

to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

{0, (3/5) * (6/5), (4/5) * (6/5)} i.e. (0, 18/25, 24/25)

(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

Given, x + y + z = 1    ………….1

The direction ratios of the normal are 1, 1, and 1.

So, √(12 + 12 + 12) = √(1 + 1 + 1) = √3

Dividing both sides of equation 1 by √3, we obtain

x/√3 + y/√3 + z/√3 = 1/√3

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal

to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

{(1/√3) * (1/√3), (1/√3) * (1/√3), (1/√3) * (1/√3)} i.e. (1/3, 1/3, 1/3)

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

Given, 5y + 8 = 0

=> 0x − 5y + 0z = 8       ...........1

The direction ratios of the normal are 0, −5, and 0.

So, √{12 + (-5)2 + 12} = √(0 + 25 + 0) = √25 = 5

Dividing both sides of equation 1 by 5, we get

-y = 8/5

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal

to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

{0, (-1) * (8/5), 0} i.e. (0, -8/5, 0)

Question 5:

Find the vector and Cartesian equation of the planes

(a) that passes through the point (1, 0, −2) and the normal to the plane is i + j - k.

(b) that passes through the point (1, 4, 6) and the normal vector to the plane is i – 2j + k

Answer:

(a) The position vector of point (1, 0, −2) is a = i – 2k

The normal vector N perpendicular to the plane is N = i + j - k

The vector equation of the plane is given by,

      (r - a).N = 0

=> [r – (i – 2k)].(i + j - k) = 0    ……………1

r is the position vector of any point P(x, y, z) in the plane.

So, r = xi + yj + zk

Therefore, equation 1 becomes

      [(xi + yj + zk) – (i – 2k)].(i + j - k) = 0

=> [(x – 1)i + yj + (z + 2)k] .(i + j - k) = 0

=> (x – 1) + y - (z + 2) = 0

=> x + y – z – 3 = 0

=> x + y – z = 3

This is the Cartesian equation of the required plane.

(b) The position vector of the point (1, 4, 6) is a = i + 4j + 6k

The normal vector perpendicular to the plane is N = i – 2j + k

The vector equation of the plane is given by,

      (r - a).N = 0

=> [r – (i + 4j + 6k)].(i - 2j + k) = 0    ……………1

r is the position vector of any point P(x, y, z) in the plane.

So, r = xi + yj + zk

Therefore, equation 1 becomes

      [(xi + yj + zk) – (i + 4j + 6k)].(i - 2j + k) = 0

=> [(x – 1)i + (y - 4)j + (z - 6)k] .(i - 2j + k) = 0

=> (x – 1) – 2(y - 4) + (z - 6) = 0

=> x - 2y + z + 1 = 0

This is the Cartesian equation of the required plane.

 

 

Question 6:

Find the equation of the planes that passes through three points.

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)                

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

Answer:

(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).

Now,    1       1       -1

             6       4        -5    = (12 - 10) – (18 - 20) – (-12 + 16) = 2 + 2 – 4 = 0

            -4      -2        3

Since A, B, C are collinear points, there will be infinite number of planes passing through the

given points.

(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).

Now,    1       1         0

             1       2         1    = (-2 - -2) – (-2 + 2) + 0 = -4 – 4 = -8 ≠ 0

            -2       2       -1

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points (x1, y1, z1), (x2, y2, z2) and

(x3, y3, z2) is

    x – x1         y – y1          z – z1

   x2 – x1        y2 – y1         z2 – z1     = 0

   x3 – x1        y3 – y1         z3 – z1

 

 

=>   x - 1      y - 1     z   

        0            1         1    = 0

       -3            1       -1

=> -2(x - 1) – 3(y – 1) + 3z = 0

=> -2x – 3y + 3z + 2 + 3 = 0

=> -2x – 3y + 3z = 5

=> 2x + 3y – 3z = 5

This is the Cartesian equation of the required plane.

Question 7:

Find the intercepts cut off by the plane 2x + y – z = 5

Answer:

Given, 2x + y – z = 5      ……………..1

Dividing both sides of equation 1 by 5, we obtain

       2x/5 + y/5 – z/5 = 5/5

=> x/(5/2) + y/5 + z/(-5) = 1      ………….2

It is known that the equation of a plane in intercept form is x/a + y/b + z/c = 0, where a, b, c

are the intercepts cut off by the plane at x, y, and z axes respectively.

Therefore, for the given equation,

a = 5/2, b = 5 and c = -5

Thus, the intercepts cut off by the plane are 5/2, 5 and -5

Question 8:

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

 

Answer:

The equation of the plane ZOX is y = 0

Any plane parallel to it is of the form, y = a

Since the y-intercept of the plane is 3,

So, a = 3

Thus, the equation of the required plane is y = 3

Question 9:

Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and     x + y + z – 2 = 4 and the point (2, 2, 1).

Answer:

The equation of any plane through the intersection of the planes 3x − y + 2z − 4 = 0 and

x + y + z − 2 = 0, is

(3x − y + 2z – 4) + λ(x + y + z – 2) = 0, where λ є R   …………..1

The plane passes through the point (2, 2, 1).

Therefore, this point will satisfy the equation 1.

So, (3 * 2 − 2 + 2 * 1 – 4) + λ(2 + 2 + 1 – 2) = 0

=> 2 + 3λ = 0

=> λ = -2/3

Put λ = -2/3 in equation 1, we get

      (3x − y + 2z – 4) + (-2/3)(x + y + z – 2) = 0

=> 3(3x − y + 2z – 4) - 2(x + y + z – 2) = 0

=> 9x − 3y + 6z – 12 - 2x - 2y - 2z + 4 = 0

=> 7x – 5y + 4z – 8 = 0

This is the required equation of the plane.

Question 10:

Find the vector equation of the plane passing through the intersection of the planes                   r.(2i + 2j – 3k) = 7, r.(2i + 5j + 3k) = 9 and through the point (2, 1, 3).

Answer:

Given, equations of plane are

r.(2i + 2j – 3k) = 7   ………….1

r.(2i + 5j + 3k) = 9   ………….2

The equation of any plane through the intersection of the planes given in equations 1 and 2 is

given by,

      [r.(2i + 2j – 3k) - 7] + λ[r.(2i + 5j + 3k) - 9] = 0, where λ є R

=> r.[(2i + 2j – 3k) + λ[r.(2i + 5j + 3k)] = 9λ + 7

=> r.[(2 + 2λ)i + (2 + 5λ)j + (3λ - 3)k] = 9λ + 7      ……………….3

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

r = 2i + j + 3k

Substitute the value of r in the equation 3, we get

=> (2i + j + 3k).[(2 + 2λ)i + (2 + 5λ)j + (3λ - 3)k] = 9λ + 7

=> 2(2 + 2λ) + (2 + 5λ) + 3(3λ - 3) = 9λ + 7

=> 4 + 4λ + 2 + 5λ + 9λ - 9 = 9λ + 7

=> 9λ = 10

=> λ = 10/9

Substituting λ = 10/9 in equation 3, we obtain

=> r.[(2 + 2 * 10/9)i + (2 + 5 * 10/9)j + (3 * 10/9 - 3)k] = 9 * 10/9 + 7

=> r.[(2 + 20/9)i + (2 + 50/9)j + (30/9 - 3)k] = 10 + 7

=> r.[38i/9 + 68j/9 + 3k/9] = 17

=> r.(38i + 68j + 3k) = 153

This is the vector equation of the required plane.

Question 11:

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and      2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0

Answer:

The equation of the plane through the intersection of the planes x + y + z = 1 and

2x + 3y + 4z = 5 is

      (x + y + z - 1) + λ(2x + 3y + 4z - 5) = 0

=> (2λ + 1)x + (3λ + 1)y + (4λ + 1)z – (5λ + 1)= 0     ……………….1

The direction ratios, a1, b1, c1 of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).

The plane in equation 1 is perpendicular to x – y + z = 0

Its direction ratios, a2, b2, c2 are 1, −1, and 1.

Since the planes are perpendicular,

=> a1a2 + b1b2 + c1c2 = 0

=> (2λ + 1) - (3λ + 1) + (4λ + 1) = 0

=> 3λ + 1 = 0

=> λ = -1/3

Put λ = -1/3 in equation 1, we get

=> (2 * -1/3 + 1)x + (3 * -1/3 + 1)y + (4 * -1/3 + 1)z + (5 * -1/3 + 1) = 0

=> (-2/3 + 1)x + (-1 + 1)y + (-4/3 + 1)z + (-5/3 + 1) = 0

=> x/3 –z/3 + 2/3 = 0

=> x – z + 2 = 0

This is the required equation of the plane.

Question 12:

Find the angle between the planes whose vector equations are                                                       r.(2i + 2j – 3k) = 5 and r.(3i - 3j + 5k) = 3

Answer:

The equations of the given planes are:

r.(2i + 2j – 3k) = 5 and r.(3i - 3j + 5k) = 3

It is known that if n1 and n2 are normal to the planes r. n1 = d1 and r. n2 = d2

then the angle between them, Q, is given by,

cos Q = |(n1 . n2)/(|n1|n2|)|     ………………1

Here, n1 = (2i + 2j – 3k)

And n2 = (3i - 3j + 5k)

Now, n1 . n2 = (2i + 2j – 3k).(3i - 3j + 5k) = 2 * 3 – 2 * 3 – 3 * 5 = 6 – 6 – 15 = -15

|n1| = √{(2)2 + (2)2 + (-3)2} = √(4 + 4 + 9) = √17

|n2| = √{(3)2 + (-3)2 + (5)2} = √(9 + 9 + 25) = √43

From equation 1, we get

      cos Q = |-15/(√17 * √43)|

=> cos Q = 15/√731

=> Q = cos-1(15/√731)

Question 13:

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

(b) 2x + y + 3z - 2 = 0 and x – 2y + 5 = 0

(c) 2x - 2y + 4z + 5 = 0 and 3x – 3y + 6z - 1 = 0

(d) 2x - y + 3z - 1 = 0 and 2x – y + 3z + 3 = 0

(e) 4x + 8y + z - 8 = 0 and y + z - 4 = 0

Answer:

The direction ratios of normal to the plane L1: a1x + b1y + c1z = 0 are a1, b1, c1 and

L2: a2x + b2y + c2z = 0 are a2, b2, c2

L1 || L2 if a1/a2 = b1/b2 = c1/c2 

L1 Ʇ L2 if a1a2 + b1b2 + c1c2 = 0

The angle between L1 and L2 is given by,

Q = cos-1|(a1a2 + b1b2 + c1c2)/{√(a12 + b12 + c12). √(a22 + b22 + c22)}|

(a) The equations of the planes are 2x + y + 3z - 2 = 0 and x – 2y + 5 = 0

Here, a1 = 7, b1 = 5, c1 = 6 and a2 = 3, b2 = -1, c2 = -10

a1a2 + b1b2 + c1c2 = 7 * 3 + 5 * (-1) + 6 * (-10) = 21 – 5 – 60 = -44 ≠ 0

Therefore, the given planes are not perpendicular to each other.

a1a2 = 7/3, b1b2 = 5/(-1) = -5, c1c2 = 6/(-10) = -3/5

It can be seen that a1a2 ≠ b1b2 ≠ c1c2

Therefore, the given planes are not parallel to each other.

The angle between them is given by,

      Q = cos-1|{7 * 3 + 5 * (-1) + 6 * (-10)}/{√(72 + 52 + 62). √{32 + (-1)2 + (-10)2}|

=> Q = cos-1|(21 – 5 - 60)/{√(110). √(110)|

=> Q = cos-1(44/110)

=> Q = cos-1(2/5)

(b) The equations of the planes are 7x + 5y + 6z + 30 = 0 and 3x − y − 10z + 4 = 0

Here, a1 = 2, b1 = 1, c1 = 3 and a2 = 1, b2 = -2, c2 = 0

Now, a1a2 + b1b2 + c1c2 = 2 * 1 + 1 * (-2) + 3 * 0 = 2 – 2 + 0 = 0

Therefore, the given planes are perpendicular to each other.

(c) The equations of the planes are 2x - 2y + 4z + 5 = 0 and 3x – 3y + 6z - 1 = 0

Here, a1 = 2, b1 = -2, c1 = 4 and a2 = 3, b2 = -3, c2 = 6

Now, a1a2 + b1b2 + c1c2 = 2 * 3 + (-2) * (-3) + 4 * 6 = 6 + 6 + 24 = 36 ≠ 0

Therefore, the given planes are not perpendicular to each other.

a1a2 = 2/3, b1b2 = -2/(-3) = 2/3, c1c2 = 4/6 = 2/3

It can be seen that a1a2 = b1b2 = c1c2

Therefore, the given planes are parallel to each other.

(d) The equations of the planes are 2x - y + 3z - 1 = 0 and 2x – y + 3z + 3 = 0

Here, a1 = 2, b1 = -1, c1 = 3 and a2 = 2, b2 = -1, c2 = 3

Here, a1a2 = 2/2 = 1, b1b2 = -1/(-1) = 1, c1c2 = 3/3 = 1

It can be seen that a1a2 = b1b2 = c1c2

Therefore, the given planes are parallel to each other.

(e) The equations of the planes are 4x + 8y + z - 8 = 0 and y + z - 4 = 0

Here, a1 = 4, b1 = 8, c1 = 1 and a2 = 0, b2 = 1, c2 = 1

Now, a1a2 + b1b2 + c1c2 = 4 * 0 + 8 * 1 + 1 * 1 = 0 + 8 + 1 = 9 ≠ 0

Therefore, the given planes are not perpendicular to each other.

Again, a1a2 = 4/0, b1b2 = 8/1 = 8, c1c2 = 1/1 = 15

It can be seen that a1a2 ≠ b1b2 ≠ c1c2

Therefore, the given planes are not parallel to each other.

The angle between them is given by,

      Q = cos-1|{4 * 0 + 8 * 1 + 1 * 1}/{√(42 + 82 + 12). √(02 + 12 + 12)|

=> Q = cos-1|(8 + 1)/{√(81). √2}|

=> Q = cos-1|(8 + 1)/9√2|

=> Q = cos-1(9/9√2)

=> Q = cos-1(1/√2)

=> Q = 450

Question 14:

In the following cases, find the distance of each of the given points from the corresponding given plane.

      Point                                 Plane

 (a) (0, 0, 0)                    3x – 4y + 12z = 3

(b) (3, −2, 1)                   2x – y + 2z + 3 = 0

(c) (2, 3, −5)                    x + 2y – 2z = 9

(d) (−6, 0, 0)                   2x – 3y + 6z – 2 = 0

Answer:

It is known that the distance between a point, p(x1, y1, z1), and a plane, Ax + By + Cz = D, is

given by,

d = |(Ax1 + By1 + Cz1 – D)/√(A2 + B2 + C2)|    ………………..1

(a) The given point is (0, 0, 0) and the plane is 3x – 4y + 12z = 3

Now, d = |(3 * 0 – 4 * 0 + 12 * 0 - 3)/ √{32 + (-4)2 + (12)2}|

=> d = |-3/√(9 + 16 + 144)|

=> d = 3/√169

=> d = 3/13

(b) The given point is (3, − 2, 1) and the plane is 2x – y + 2z + 3 = 0

Now, d = |(2 * 3 + 2 * 1 + 1 * 2 + 3)/ √{22 + (-1)2 + (2)2}|

=> d = |(6 + 2 + 2 + 3)/√(4 + 1 + 4)|

=> d = 13/√9

=> d = 13/3

(c) The given point is (2, 3, −5) and the plane is x + 2y – 2z = 9

Now, d = |(2 * 1 + 3 * 2 + 5 * 2 - 9)/ √{12 + 22 + (-2)2}|

=> d = |(2 + 6 + 10 - 9)/√(1 + 4 + 4)|

=> d = 9/√9

=> d = 3/3

=> d = 1

(d) The given point is (−6, 0, 0) and the plane is 2x – 3y + 6z – 2 = 0

Now, d = |(-2 * 6 - 3 * 0 + 6 * 0 - 2)/ √{22 + (-3)2 + 62}|

=> d = |(-12 + 0 + 0 - 2)/√(4 + 9 + 36)|

=> d = 14/√49

=> d = 14/7

=> d = 2

 

 

 

 

 

 

 

 

                                                     Miscellaneous Exercise on Chapter 11

Question 1:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1), (4, 3, −1).

Answer:

Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).

Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).

The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and (−1 + 1) = 0

OA is perpendicular to BC, if a1a2 + b1b2 + c1c2 = 0

Now, a1a2 + b1b2 + c1c2 = 2 * 1 + 1 * (−2) + 1 * 0 = 2 − 2 = 0

Thus, OA is perpendicular to BC.

Question 2:

 

 

Question 3:

Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.

Answer:

The angle Q between the lines with direction cosines, a, b, c and b − c, c − a, a − b, is given by,

      cos Q = |{a(b - c) + b(c - a) + c(a - b)}/[√(a2 + b2 + c2) + √{(b - c)2 + (c - a)2 + (a - b)2)}]|

=> cos Q = 0

=> cos Q = cos 90°

=> Q = 90°

Thus, the angle between the lines is 90°.

 

Question 4:

Find the equation of a line parallel to x-axis and passing through the origin.

Answer:

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where a є R.

Direction ratios of OA are (a − 0) = a, 0, 0

The equation of OA is given by,

      (x - 0)/a = (y - 0)/0 = (z - 0)/0

=> x/1 = y/0 = z/0

Thus, the equation of line parallel to x-axis and passing through origin is x/1 = y/0 = z/0

Question 5:

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer:

The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and (2, 9, 2) respectively.

The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4

The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8

It can be seen that, a1a2 = b1b2 = c1c2 = 1/2

Therefore, AB is parallel to CD.

Thus, the angle between AB and CD is either 0° or 180°.

Question 6:

If the lines (x - 1)/(-3) = (y - 2)/3k = (z - 3)/2 and (x - 1)/3k = (y - 1)/5 = (z - 6)/(-5) are perpendicular, find the value of k.

 

Answer:

The direction of ratios of the lines

(x - 1)/(-3) = (y - 2)/3k = (z - 3)/2 and (x - 1)/3k = (y - 1)/5 = (z - 6)/(-5)

are −3, 2k, 2 and 3k, 1, −5 respectively.

It is known that two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular, if

a1a2 + b1b2 + c1c2 = 0

=> -3 * 3k + 2k * 1 + 2 * (-5) = 0

=> -9k + 2k – 10 = 0

=> 7k = -10

=> k = -10/7

Therefore, for k = -10/7, the given lines are perpendicular to each other.

Question 7:

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane r.(i + 2j – 5k) + 9 = 0

Answer:

The position vector of the point (1, 2, 3) is r1 = i + 2j + 3k

The direction ratios of the normal to the plane r.(i + 2j – 5k) + 9 = 0 are 1, 2, and −5 and the

normal vector is N = i + 2j – 5k

The equation of a line passing through a point and perpendicular to the given plane is given by,

      l = r + λN, λ є R

=> l = (i + 2j + 3k) + λ(i + 2j - 5k)

Question 8:

Find the equation of the plane passing through (a, b, c) and parallel to the plane r.(i + j + k) = 2

 

Answer:

Any plane parallel to the plane r1.(i + j + k) = 2 is

r.(i + j + k) = λ    …………1

The plane passes through the point (a, b, c). Therefore, the position vector r of this point is

 r = ai + bj + ck

Therefore, equation 1 becomes

(ai + bj + ck).(i + j + k) = λ

=> a + b + c = λ

Substituting λ = a + b + c in equation 1, we obtain

r.(i + j + k) = a + b + c   …………….2

This is the vector equation of the required plane.

Substituting r = xi + yj + zk in equation 2, we get

      (xi + yj + zk).(i + j + k) = a + b + c

=> x + y + z = a + b + c

Question 9:

 

 

Question 10:

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the              YZ-plane.

Answer:

It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,y2, z2) is

(x – x1)/(x2 – x1) = (y – y1)/(y2 – y1) = (z – z1)/(z2 – z1)

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

      (x – 5)/(3 – 5) = (y – 1)/(4 – 1) = (z – 6)/(1 – 6)

=> (x – 5)/(-2) = (y – 1)/3 = (z – 6)/(-5) = k (say)

=> x = 5 – 2k, y = 3k + 1, z = 6 – 5k

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

The equation of YZ-plane is x = 0

Since the line passes through YZ-plane,

=> 5 – 2k = 0

=> k = 5/2

3k + 1 = 3 * (5/2) + 1 = 15/2 + 1 = 17/2

6 – 5k = 6 – 5 * (5/2) = 6 – 25/2 = -13/2

Therefore, the required point is (0, 17/2, -13/2).

Question 11:

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the              ZX-plane.

Answer:

It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,y2, z2) is

(x – x1)/(x2 – x1) = (y – y1)/(y2 – y1) = (z – z1)/(z2 – z1)

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

      (x – 5)/(3 – 5) = (y – 1)/(4 – 1) = (z – 6)/(1 – 6)

=> (x – 5)/(-2) = (y – 1)/3 = (z – 6)/(-5) = k (say)

=> x = 5 – 2k, y = 3k + 1, z = 6 – 5k

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

The equation of ZX-plane is y = 0

Since the line passes through ZX-plane,

=> 3k + 1 = 0

=> k = -1/3

5 – 2k = 5 - 2 * (-1/3) = 5 + 2/3 = 17/3

6 – 5k = 6 – 5 * (-1/3) = 6 + 5/3 = 23/3

Therefore, the required point is (17/3, 0, 23/3).

Question 12:

Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7.

Answer:

It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,y2, z2) is

(x – x1)/(x2 – x1) = (y – y1)/(y2 – y1) = (z – z1)/(z2 – z1)

Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is given by,

      (x – 3)/(2 – 3) = (y + 4)/(-3 + 4) = (z + 5)/(1 + 5)

=> (x – 3)/(-1) = (y + 4)/1 = (z + 5)/6 = k (say)

=> x = 3 – k, y = k - 4, z = 6k – 5

Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5).

This point lies on the plane, 2x + y + z = 7

=> 2(3 − k) + (k − 4) + (6k − 5) = 7

=> 6 – 2k + k – 4 + 6k – 5 = 7

=> 5k - 3 = 7

=> 5k = 10

=> k = 2

Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 * 2 − 5) i.e., (1, −2, 7).

 

Question 13:

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answer

The equation of the plane passing through the point (−1, 3, 2) is

a(x + 1) + b(y − 3) + c (z − 2) = 0     ......... 1

where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are

perpendicular, if a1a2 + b1b2 + c1c2 = 0

Plane 1 is perpendicular to the plane x + 2y + 3z = 5

So, a * 1 + b * 2 + c * 3 = 0

=> a + 2b + 3c = 0   ……………..2

Also, plane 1 is perpendicular to the plane 3x + 3y + z = 0

So, a * 3 + b * 3 + c * 1 = 0

=> 3a + 3b + c = 0   ……………..3

From equations 2 and 3, we get

       a/(2 * 1 – 3 * 3) = b/(3 * 3 – 1 * 1) = c/(1 * 3 – 2 * 3)

=> a/(2 - 9) = b/(9 - 1) = c/(3 - 6)

=> a/(-7) = b/8 = c/(-3) = k (say)

=> a = -7k, b = 8k, c = -3k

Substituting the values of a, b, and c in equation 1, we get

      -7k(x + 1) + 8k(y - 3) – 3k(z - 2) = 0

=> -7(x + 1) + 8(y - 3) – 3(z - 2) = 0

=> -7x – 7 + 8y – 24 – 3z + 6 = 0

=> 7x – 8y + 3z + 25 = 0

This is the required equation of the plane.

Question 14:

 

 

Question 15:

 

 

Question 16:

If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

Answer

The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1, y1, z1) is

a(x – x1) + b(y – y1) + c(z – z1) = 0 where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

      1(x – 1) + 2(y – 2) - 3(z + 3) = 0

=> x – 1 + 2y – 4 – 3z – 9 = 0

=> x + 2y – 3z – 14 = 0

Question 17:

 

 

Question 18:

 

 

Question 19:

 

 

Question 20:

 

 

Question 21:

Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the origin, then 1/a2 + 1/b2 + 1/c2 = 1/p2

Answer:

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

x/a + y/b + z/c = 1    …………..1

The distance (p) of the plane from the origin is given by,

      p = |(0/a + 0/b + 0/c - 1)/√{(1/a)2 + (1/b)2 + (1/c)2}|

=> p = 1/√{(1/a)2 + (1/b)2 + (1/c)2}

=> p2 = 1/{(1/a)2 + (1/b)2 + (1/c)2}

=> 1/p2 = 1/a2 + 1/b2 + 1/c2

Question 22:

Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is                                       (A) 2 units                      (B) 4 units                 (C) 8 units                      (D) 2/√29 units

 

Answer:

The equations of the planes are:

2x + 3y + 4z = 4    …………..1

4x + 6y + 8z = 12

=> 2x + 3y + 4z = 6    …………..2

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, ax + by + cz = d1 and

ax + by + cz = d2 is given by,

      D = |(d2 – d1)/√(a2 + b2 + c2)|

=> D = |(6 – 4)/√(22 + 32 + 42)|

=> D = |2/√(4 + 9 + 16)|

=> D = 2/√29

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is option D.

Question 23:

The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are

(A) Perpendicular          (B) Parallel          (C) intersect y-axis         (D) passes through (0, 0, 5/4)

Answer:

The equations of the planes are

2x − y + 4z = 5     ............1

5x − 2.5y + 10z = 6   .......2

It can be seen that,

a1a2 = 2/5

b1b2 = -1/(-2.5) = 10/25 = 2/5

c1c2 = 4/10 = 2/5

So, a1a2 = b1b2 = c1c2

Therefore, the given planes are parallel.

Hence, the correct answer is option B.

 

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