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VITEEE 2021 Paper

Section: Mathematics

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Question : 53 of 80

Marks: +1, -0

The shortest distance between the lines

x=y+2=6z-6

x+1=2y=-12z

is

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

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,

\vec{a}_1 = -2 \hat{j} + \hat{k}, \vec{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}

Shortest distance

\frac{ \left| (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \right| }{ \left| \vec{b}_1 \times \vec{b}_2 \right| }

\frac{ \left| (-\hat{i} + 2\hat{j} - \hat{k}) \cdot (-12\hat{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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