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3D Geometry

11.3. Angle Between Two Lines.

Q.1. Find the angle between the pair of lines given by
r→ = (2i – 5j + k) + λ(3i + 2j + 6k) and r→ = 7i – 6j – 6k + μ(i + 2j + 2k).

Solution :

We have, r→ = 2i – 5j + k + λ(3i + 2j + 6k) and r→ = 7i – 6j – 6k + μ(i + 2j + 2k)
Comparing by r→ = a→ + λ b→,
we have a→1 = 2i – 5j + k, b→1 = 3i + 2j + 6k,
a→2 = 7i – 6j – 6k, b→2 = i + 2j + 2k.
Angle between two lines θ is given by
cos θ = |(b→₁.b→₂)/|(b→₁||b→₂|)|
= |{(3i + 2j + 6k).(i + 2j + 2k)}/{√(9 + 4 + 36)√(1 + 4 + 4)}|
= |(3 + 4 + 12)/{√(49)√(9)}|
= 19/21
Therefore, θ = cos –1(19/21). [Ans.]

11.4. Shortest Distance Between Two Lines.

Q.1. The vector equations of two lines are :
r→ = i + 2j + 3k + λ(i – 3j + 2k ) and
r→ = 4i + 5j + 6k + μ(2i + 3j + k).
Find the shortest distance between the above lines.

Solution :

Fig.
We have, l1: r→ = (i + 2j + 3k) + λ(i – 3j + 2k) ----------------- (1)
l2: r→ = (4i + 5j + 6k) + μ(2i + 3j + k) ---------------- (2)
(1) passes through A(1, 2, 3) and is parallel to the vector i – 3j + 2k, (2) passes through B(4, 5, 6) and is parallel to the vector 2i + 3j + k.
The equations (1) and (2) can be written as
r→ = (1 + λ)i + (2 – 3λ)j + (3 + 2λ)k
r→ = (4 + 2μ)i + (5 + 3μ)j + (6 + μ)k.
The point P(1 + λ, 2 – 3λ, 3 + 2λ) lie on (1) and Q(4 + 2μ, 5 + 3μ, 6 + μ) lie on (2)
Therefore, PQ→ = OQ→ – OP→
= (4 + 2μ – 1 – λ)i + (5 + 3μ – 2 + 3λ)j + (6 + μ – 3 – 2λ)k
= (2μ – λ + 3)i + (3μ + 3λ + 3)j + (μ – 2λ + 3)k.
If PQ→ is taken as the shortest distance between (1) and (2), then PQ is perpendicular to both (1) and (2).
Hence, (2μ – λ + 3).1 + (3μ + 3λ + 3).(–3) + (μ – 2λ + 3).2 = 0
=> 5μ + 14λ + 0 = 0 ----- (3)
and (2μ – λ + 3). 2 + (3μ + 3λ + 3). 3 + (μ – 2λ + 3). 1 = 0
=> 14 μ + 5 λ + 18 = 0 ----- (4)
Solving (3) and (4) by cross multiplication, we get
μ/–252 = λ/90 = 1/(– 25 + 196)
Or, μ = – 28/19, λ = 10/9.
Therefore, PQ→ = – 9/19 i + 3/19 j + 9/19 k
Thus, S. D. = |PQ→| = √{(– 9/19)² + (3/19)² + (9/19)²}
= (√171)/19
= 3/√19 units. [Ans.]

Paper By Mr. M.P.Keshari
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