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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 24 of 78
Marks: +1, -0
Find the shortest distance between the lines whose vector equations are r⃗=(1−t)i^+(t−2)j^+(3−2t)k^\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k} and r⃗=(s−1)i^+(2s−1)j^−(2s+1)k^\vec{r} = (s-1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}
Solution:  
From the given equations, we get
r⃗=i^−2j^+3k^\vec{r} = \hat{i} -2\hat{j} + 3\hat{k} +t(−i^+j^−2k^)+t(-\hat{i} + \hat{j} -2\hat{k}) ...(1)
r⃗=i^−j^−k^\vec{r} = \hat{i} - \hat{j} - \hat{k} +s(i^+2j^−2k^)+s(\hat{i} +2\hat{j} -2\hat{k}) ...(2)
Equations (1) and (2) are of the form r⃗=a1⃗+tb1⃗\vec{r} = \vec{a_1} + \vec{tb_1} and r⃗=a2⃗+sb2⃗\vec{r} = \vec{a_2} + \vec{sb_2}
where a1⃗=i^−2j^+3k^\vec{a_1} = \hat{i} -2\hat{j} + 3\hat{k} , a2⃗=i^−j^−k^\vec{a_2} = \hat{i} - \hat{j} - \hat{k}
and b1⃗=−i^+j^−2k^\vec{b_1} = -\hat{i} + \hat{j} -2\hat{k} , b2=i^+2j^−2k^b_2 = \hat{i} +2\hat{j} -2\hat{k}
Here a2⃗−a1⃗=j^−4k^\vec{a_2} - \vec{a_1} = \hat{j} - \hat{4k}
Also, b1⃗×b2⃗=∣i^j^k^−11−212−2∣\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} =2i^−4j^−3k^= \hat{2i} - \hat{4j} - \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})}{|\vec{b_1} \times \vec{b_2}|} \right|
=∣(2i^−4j^−3k^)⋅(j^−4k^)22+(−4)2+(−3)2∣= \left| \frac{(\hat{2i} - \hat{4j} - \hat{3k}) \cdot (\hat{j} - \hat{4k})}{\sqrt{2^2 + (-4)^2 + (-3)^2}} \right|
=829= \frac{8}{\sqrt{29}}units
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