JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 8)
The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :
3
$$\sqrt{14}$$
$$\sqrt6$$
$$2\sqrt3$$
Explanation
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$\overrightarrow{A P}=7 \hat{i}+4 \hat{j}-5 \hat{k} \Rightarrow|\overrightarrow{A P}|=\sqrt{49+16+25}=\sqrt{90} A N$
$=$ projection of $\overrightarrow{A P}$ on $\vec{b}=\overrightarrow{A P} \cdot \vec{b}=\frac{21+12+5}{\sqrt{19}}=\frac{38}{\sqrt{19}}$
$(P N)^{2}=(A P)^{2}-(A N)^{2}=90-76=14 \Rightarrow P N=\sqrt{14}$
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