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

3D Geometry

Alternative Method.

Let the equations (1) and (2) be written as, r→ = a→ + λb→ and
r→ = c→ + μd→.
Where, a→ = i + 2j + 3k, b→ = 2i + 3j + k,
c→ = 4i + 5j + 6k and d→ = 2i + 3j + k.
Therefore, a→ – c→ = – 3i – 3j – 3k and
|i j k|
b→×d→ = |1 –3 2 |
|2 3 1|
= – 9i + 3j + 9k
Therefore, (a→ – c→).(b→×d→) = (–3i – 3j – 3k).(– 9i + 3j + 9k)
=(– 3)(– 9) – 3(3) – 3(9)
= 27 – 9 – 27
= – 9.
and |b→×d→| = √[(0 – 9)² + (3)² + (9)²]
= √(81 + 9 + 82)
= √171
= 3√19.
Therefore, S.D. = |[(a→ – c→).(b→×d→)]/[|b→ – d→|]|
= 9/3√19
= 3/√19. [Ans.]

Q.2. Find the shortest distance between the following lines :
(x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and
(x + 1)/7 = (y + 1)/–6 = (z + 1)/1.

Solution :

We have,
(x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and
(x + 1)/7 = (y + 1)/–6 = (z + 1)/1
Which in vector form are :
r→ = (3i + 5j + 7k) + λ(i – 2j + k) -------------------- (1)
r→ = (– i – j – k) + μ(7i – 6j + k) --------------------- (2)
Here, a→1 = 3i + 5j + 7k, b→1 = i – 2j + k, a→2 = – i – j – k and b→2 = 7i – 6j + k.
Shortest distance, S.D. = |(b→1×b→2).(a→2 – a→1)/|b→1×b→2||.
|i j k|
b→1×b→2 = |1 –2 1|
|7 –6 1|
= i(–2 + 6) – j(1 – 7) + k(–6 + 14)
= 4i + 6j + 8k.
|b→1×b→2| = √(16 + 36 + 64) = √116.
a→2 – a→1 = – i – j – k – 3i – 5j – 7k = – 4i – 6j – 8k.
Therefore, S.D. = |{(4i + 6j + 8k).(– 4i – 6j – 8k)}/√116|
= |(– 16 – 36 – 64)/√116|
= |– 116/√116|
= √116 = 2√29. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289