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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 21 of 78
Marks: +1, -0
Find the shortest distance between the lines r⃗=(i^+2j^+k^)\vec{r}=(\hat{i}+\hat{2j}+\hat{k}) +λ(i^−j^+k^)+\lambda(\hat{i}-\hat{j}+\hat{k}) and r⃗=(2i^−j^−k^)\vec{r}=(\hat{2i}-\hat{j}-\hat{k}) +μ(2i^+j^+2k^)+\mu(\hat{2i}+\hat{j}+\hat{2k})
Solution:  
Comparing the given lines by r⃗=a1⃗+λb1⃗\vec{r}=\vec{a_1}+\lambda\vec{b_1} and r⃗=a2⃗+μb2⃗.\vec{r}=\vec{a_2}+\mu\vec{b_2}.
We get a1⃗=i^+2j^+k^\vec{a_1}=\hat{i}+\hat{2j}+\hat{k} , a2⃗=2i^−j^−k^\vec{a_2}=\hat{2i}-\hat{j}-\hat{k}
and b1⃗=i^−j^+k^\vec{b_1}=\hat{i}-\hat{j}+\hat{k}, b2⃗=2i^+j^+2k^\vec{b_2}=\hat{2i}+\hat{j}+\hat{2k}
Now a2⃗−a1⃗=i^−3j^−2k^\vec{a_2}-\vec{a_1}= \hat{i}-\hat{3j}-\hat{2k}
and b1⃗×b2⃗=∣i^j^k^1−11212∣\vec{b_1}\times\vec{b_2}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} =−3i^+3k^=-\hat{3i}+\hat{3k}
Required shortest distance =∣(b1⃗×b2⃗)⋅(a2⃗−a1⃗)∣b1⃗×b2⃗∣∣=\left| \frac{(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1})}{\lvert\vec{b_1}\times\vec{b_2}\rvert} \right|
=∣(−3i^+3k^)⋅(i^−3j^−2k^)∣−3i^+3k^∣∣=\left| \frac{(-\hat{3i}+\hat{3k})\cdot(\hat{i}-\hat{3j}-\hat{2k})}{\lvert-\hat{3i}+\hat{3k}\rvert} \right|
=918=32=\frac{9}{\sqrt{18}}=\frac{3}{\sqrt{2}}units
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