Test Index

NCERT Class XII Mathematics Chapter - - Solutions

© examsnet.com
Question : 23 of 78
Marks: +1, -0
Find the shortest distance between the lines whose vector equations are r⃗=i^+2j^+3k^\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} +λ(i^−3j^+2k^)+ \lambda ( \hat{i} - 3\hat{j} + 2\hat{k} ) and r⃗=4i^+5j^+6k^\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} +μ(2i^+3j^+k^)+ \mu ( 2\hat{i} + 3\hat{j} + \hat{k} )
Solution:  
Equations of the given lines are of the form r⃗=a⃗1+λb⃗1\vec{r} = \vec{a}_1 + \lambda \vec{b}_1 and r⃗=a⃗2+λb⃗2\vec{r} = \vec{a}_2 + \lambda \vec{b}_2
where a⃗1=i^+2j^+3k^\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k} , a⃗2=4i^+5j^+6k^\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}
and b⃗1=4i^+5j^+6k^\vec{b}_1 = 4\hat{i} + 5\hat{j} + 6\hat{k} , b2=2i^+3j^+k^b_2 = 2\hat{i} + 3\hat{j} + \hat{k}
Here a⃗2−a⃗1=3i^+3j^+3k^\vec{a}_2 - \vec{a}_1 = 3\hat{i} + 3\hat{j} + 3\hat{k}
Also, b⃗1×b⃗2=∣i^j^k^1−32231∣\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} =−9i^+3j^+9k^= -9\hat{i} + 3\hat{j} + 9\hat{k}
∴ Required shortest distance =∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣= \left| \frac{ (\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) }{ |\vec{b}_1 \times \vec{b}_2| } \right|
=∣−9×3+3×3+9×3(−9)2+32+92∣= \left| \frac{ -9 \times 3 + 3 \times 3 + 9 \times 3 }{ \sqrt{ (-9)^2 + 3^2 + 9^2 } } \right|
=∣9319∣=319= \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}}
© examsnet.com
Go to Question:
Prev Question
Next Question