Class 12 Maths Three Dimensional Geometry Angle between two lines (in terms of Direction cosines)

Angle between two lines (in terms of Direction cosines)

Considering 2 line vectorsP and Q and the direction cosines of P (l1, m1, n1) and of

Q(l2, m2, n2). Magnitude of P =r1 and magnitude of Q = r2.

Therefore P->=r1(l1î+m1ĵ +n1k̂) and Q-> =r2(l2î+m2ĵ +n2k̂)

P->.Q->= IPIIQIcosϴ

cosϴ= (P->.Q->)/(IPIIQI)

=r1 (l1î +m1ĵ +n1k̂). r2(l2î +m2ĵ +n2k̂)/(r1√l1)2+(m1)2+(n1)2) (r2√((l2)2+(m2)2+(n2)2)

cosϴ =Il1l2+m1m2+n1n2I(using for any line (l) 2+ (m) 2+ (n) 2 =1)

Perpendicular and parallel Test between 2 line vectors

Two lines with direction ratios a1, b1, c1and a2,b2,c2.

When Perpendicularϴ=900:-

When direction ratios are given then a1a2+b1b2+c1c2=0,and when direction cosines are given then l1l2+m1m2+n1n2=0.

When Parallelϴ=0:-

(a1/a2)=(b1/b2)=(c1/c2) or (l1/l2)/(m1/m2)=(n1/n2)=0

Angle between lines when line equations are given (Vector)

Consider the equation of a line passing through a1⃗ (point vector)and parallel to b1⃗ such that r1⃗ = a1 + λ b1⃗ andanother line passing through a2⃗  (point vector)and parallel to b2⃗ such that r2⃗= a2⃗ +μ b2⃗.

From the figure we can say that b1⃗ and b2⃗ are line vectors.

Angles between lines r1⃗ and r2⃗ are same as angle between line vectors b1⃗ and b2

Problem:-

Find the angle between the following pairs of lines:

r ⃗=-++λ (3î-2ĵ +6k̂) and r ⃗=-6k̂+μ (î+2ĵ +2k̂)?

Answer:-

=3+4+12

=19

=> cosϴ = (19)/ (7x3)

=>ϴ = cos-1(19/21)

Problem:-

Find the angle between the following pairs of lines:

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

Answer:-

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

Therefore b1⃗=2î+5ĵ -3k̂andb2⃗=-î +8ĵ +4k̂

Ib1⃗I=√ ((2)2+ (5)2+ (-3)2) =√38

Ib2⃗I= √ ((-1)2+ (8)2+ (4)2) =√81 =9

b1⃗.b2⃗ = (2î+5ĵ -3k̂). (-î+8ĵ +4k̂)

=2(-1) +5x8+ (-3).4

=-2+40-12

=26

=>cosϴ= (26)/ (9√38)

=>ϴ =cos-1(26)/ (9√38))

Problem:-

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 atright angles.

Answer:-

The given equations can be written in the standard form as:

(1-x)/(3)=(7y-14)/(2p)=(z-3)/(2)and (7-7x)/(3p)=(y-5)/(1)=(6-z)/(-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.

Therefore ((-3)x(-3p)/(7))+((2p)/(1))+((2).(-5))=0

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

=>11p=70

=>p=(70/11).

Thus, the value of p is (70/11).

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