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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 68 of 78
Marks: +1, -0
Find the shortest distance between lines r⃗=6i^+2j^+2k^\vec{r}=6\hat{i}+2\hat{j}+2\hat{k} +λ(i^−2j^+2k^)+\lambda(\hat{i}-2\hat{j}+2\hat{k}) and r⃗=−4i^−k^\vec{r}=-4\hat{i}-\hat{k} +μ(3i^−2j^−2k^).+\mu(3\hat{i}-2\hat{j}-2\hat{k}).
Solution:  
Here a⃗1=6i^+2j^+2k^\vec{a}_1=6\hat{i}+2\hat{j}+2\hat{k} , a⃗2=−4i^−k^\vec{a}_2=-4\hat{i}-\hat{k}
and b⃗1=i^−2j^+2k^\vec{b}_1=\hat{i}-2\hat{j}+2\hat{k} , b⃗2=3i^−2j^−2k^\vec{b}_2=3\hat{i}-2\hat{j}-2\hat{k}
∴a⃗2−a⃗1\therefore \vec{a}_2-\vec{a}_1 =(−4i^−k^)−(6i^+2j^+2k^)=(-4\hat{i}-\hat{k})-(6\hat{i}+2\hat{j}+2\hat{k}) =−10i^−2j^−3k^=-10\hat{i}-2\hat{j}-3\hat{k}
and b⃗1×b⃗2=∣i^j^k^1−223−2−2∣\vec{b}_1\times\vec{b}_2=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} =8i^+8j^+4k^=8\hat{i}+8\hat{j}+4\hat{k}
∴∣b⃗1×b⃗2∣=12\therefore |\vec{b}_1\times\vec{b}_2|=12
∴ Shortest distance between the given lines =∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣=\left|\frac{(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)}{|\vec{b}_1\times\vec{b}_2|}\right| = 9 units.
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