CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

3D Geometry

Q.3. Find the shortest distance between the lines
l1 : r→ = i + j + λ(2i – j + k)
l2 : r→ = 2i + j – k + μ(3i – 5j + 2k).

Solution :

On comparing by r→ = a→ + λb→
a→1 = i + j ; b→1 = 2i – j + k
And by r→ = a→ + μb→
a→2 = 2i + j – k ; b→2 = 3i – 5j + 2k
a→2 – a→1 = 2i + j – k – i – j = i – k.
|i j k|
b→1×b→2 = |2 –1 1|
|3 –5 2|
= i(–2 + 5) – j(4 – 3) + k (–10 + 3)
= 3i – j – 7k.
|b→1×b→2| = √(9 + 1 + 49) = √(59).
Therefore, shortest distance between the given lines is
S. D. = |[|b→1×b→2|.|a→2 – a→1|]/|b→1×b→2|
= |(3i – j – 7k).(i – k)√(59)|
= |(3 – 0 + 7)√(59)|
= 10/√(59). [Ans.]

11.5. Equation of a Plane.

Q.1. Find the equation of the plane passing through the points (1, 2, 3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3.

Solution :

Let the equation of the plane through (1, 2, 3) be
a(x = 1) + b(y – 2) + c(z – 3) = 0 ---------------- (1)
(1) is parallel to the line
(x – 1)/2 = (y + 2)/3 = z /–3.
Therefore, a×2 + b×3 + c×(–3) = 0
Or, 2a + 3b – 3c = 0 ------------------------------------- (2)
Also, plane (1) passes through (0, – 1, 0),
Therefore, a(0 – 1) + b(– 1 – 2) + c(0 – 3) = 0
Or, a + 3b + 3c = 0 --------------------------------------- (3)
Solving (2) and (3) by cross multiplication, we get
a/9 + 9 = b/–3 – 6 = c/6 – 3
Or, a/6 = b/ – 3 = c/1
Hence, the required plane is given by
6(x – 1) – 3 (y – 2) + 1(z – 3) = 0
Or, 6x – 3y + z = 3. [Ans.]

Paper By Mr. M.P.Keshari
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Ph No. : 09434150289