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BITSAT 2023 Shift 2 Solved Paper

Section: Mathematics

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Question : 120 of 130

Marks: +1, -0

The shortest distance between the lines

x=y+

2=6 z-6

x+1=2 y=-12 z

$\;\frac{1}{2}$
2
1
$\;\frac{3}{2}$

Solution: 👈: Video Solution

The lines are

\;\frac{x}{6}=\;\frac{y+2}{6}=\;\frac{z-1}{1}

and

\;\frac{x+1}{12}=\;\frac{y}{6}=\;\frac{z}{-1}

Here,

\;\overset{\rightarrow_1}{a}=-2\hat{j}+\hat{k}, b_1+6\hat{i}+6\hat{j}+\hat{k}, \vec{a}_2=-\hat{i},

\;\vec{b}_2=12\hat{i}+6\hat{j}-\hat{k}

\;\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 6 & 1 \\ 12 & 6 & -1 \end{vmatrix} = -12\hat{i}+18\hat{j}-36\hat{k}

\;\; \text{Shortest distance} \;=\; \frac{ \left| (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 - \vec{b}_2) \right| }{ \left| \vec{b}_1 \times \vec{b}_2 \right| }

\;=\; \frac{ \left| (-i + 2\hat{j} - \hat{k}) \cdot (-12 i + 18\hat{j} - 36\hat{k}) \right| }{ \sqrt{(-12)^2 + (18)^2 + (-36)^2} }

\;=\;\frac{\left|+12+36+36\right|}{\sqrt{1764}}=\frac{84}{42}=2

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