Class 12 Maths Three Dimensional Geometry | Direction Ratios of a Line |

**Direction Ratios of a Line**

Any 3 numbers which are proportional to the direction cosines of a line are known as direction ratios of that line.

Direction ratios are denoted by (a,b and c).

Consider a vector OP^{-->}=a î +bĵ +ck̂ (Where a, b and c are direction ratios. )

=**r (cos α ****î**** + cos****β ĵ**** +**** cos ****γ k̂****) where r=√ (a ^{2}+b^{2}+c^{2})**

cosα,cosβ and cosγ are **direction cosines**.

r cosα=a,r cosβ=b, r cosγ=c are **direction ratios**.

__Relation between Direction Ratios and Direction Cosines __

Consider a line whose one point is at originand another is at P. Considering a vector OP^{-->}=a î +bĵ +ck̂ (equation(1))

=r cos αî+r cosβĵ+r cos γk̂

where r=√ (a^{2}+b^{2}+c^{2}) (equation(2))

Comparing equations (1) and (2) :-

a=(rcos α) ,b =(r cosβ) and c=(rcosγ)(equation(3))

cos α =(a/r), cosβ=(b/r) and cosγ=(c/r) (equation(4))

Squaring and adding equation(4) :-

(a^{2}/r^{2}) + (b^{2}/r^{2})+(c^{2}/r^{2}) =1

using (cos^{2} α + cos^{2} β + cos^{2} γ) =1

=>**(a ^{2}+ b^{2}+c^{2}) = r^{2}** (equation(5))

From equation(4) **cos α =(a/r)= (±) (a)/(√( a ^{2}+ b^{2}+c^{2}) **

using equation(5)

Similarly **cos**** β=(b/r) =****(±)****(b)/**** (√( a ^{2}+ b^{2}+c^{2})** and

**cos γ=(c/r) =****(±)****(c)/**** (√( a ^{2}+ b^{2}+c^{2}) **

usingequation (5)

Therefore cos α, cos β and cosγare the __direction cosines__ which are written in the form of a, b and c which are __direction ratios__.

__Relationship between direction ratios and cosines when a line passes through 2 points__

One and only one line passes through two given points.

Consider a line PQ and the coordinates of points P and Q are given as P(x_{1}, y_{1}, z_{1}) and Q(x_{2},y_{2},z_{2}).

Let l,m and n are the direction cosines of the line PQ and which makes angleα,βand γ with the x,y and z-axis respectively.

If we draw perpendiculars from the point P and Q as shown in the figure (b), to XY-plane to meet at R and S. Also drawing a perpendicular from P to QS to meet at N.

Therefore cosα= (x_{2}-x_{1})/(PQ)

Similarly cosβ =(y_{2}-y_{1})/(PQ) and cosγ =(NQ)/(PQ) =(z_{2}-z_{1})/(PQ)

Therefore the direction cosines of the line segment joining the points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) will be:-

PQ=(x_{2}-x_{1})/(PQ) ,(y_{2}-y_{1})/(PQ) +(z_{2}-z_{1})/(PQ) equation(1)

Where **PQ=****√(**** x _{2}-x_{1})^{2}+( y_{2}-y_{1})^{2}+( z_{2}-z_{1})^{2}**

** Note:- **The direction ratios of the line segment joining point P(x

Q(x_{2}, y_{2}, z_{2}) can be considered as – (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1} or x_{1} – x_{2}, y_{1} – y_{2}, z_{1} – z_{2})

** Problem:**- If a line makes an angle of 90°, 135

** Answer:- **Let l, m and be direction cosines of the line. Then l= cos90° = 0, m=cos 135

Therefore Direction cosines: [0,-(1/√2),(1/√2)]

** Problem:- **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

l^{2}+m^{2}+n^{2} =1

=>cos^{2} α + cos^{2} β + cos^{2} γ) =1

=>3cos^{2} α =1

=>cos^{2} α = (1/3)

=>cosα = (±) (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).

** Problem:- **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})

= i.e. (-18)/ (22), (12)/ (22) and (-4)/ (22)

= (-9/11), (6/11) and (-2)/ (11)

Thus, the direction cosines are (-9/11), (6/11) and (-2)/ (11).

** Problem:- **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, (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}),are given by, (x_{2} − x_{1}), (y_{2} − y_{1}), and (z_{2} − z_{1}).

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 areproportional.

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

** Problem:- **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+32)

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

= (-4)/ (2√ (17), (-4)/(2√ (17)), (-6)/(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}),

= (-5)/ ((-8)^{2}+ (10)^{2}+ (2)^{2}),

= (2)/ ((-8)^{2}+ (10)^{2}+ (2)^{2})

=(-8)/ (2√ (42)), (-10)/ (2√ (42)), (2)/2√42))

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