JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 11)

The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :
$$2 \sqrt{29}$$
1
$$\sqrt{\frac{37}{29}}$$
$$\frac{\sqrt{29}}{2}$$

Explanation

$${L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}$$

Any point on it $${\overrightarrow a _1}( - 7,6,0)$$ and L1 is parallel to $${\overrightarrow b _1}( - 6,7,1)$$

$${L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}$$

Any point on it $${\overrightarrow a _2}(7,2,6)$$ and L2 is parallel to $${\overrightarrow b _2}( - 2,1,1)$$

Shortest distance between L1 and L2

$$ = \left| {{{({{\overrightarrow a }_2} - {{\overrightarrow a }_1})\,.\,({{\overrightarrow b }_1} \times {{\overrightarrow b }_2})} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}} \right| = \left| {{{( - 14,4, - 6)\,.\,(3,2,4)} \over {\sqrt {9 + 4 + 16} }}} \right|$$

$$ = 2\sqrt {29} $$.

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