JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 7)
The distance, of the point $(7,-2,11)$ from the line
$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :
$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :
12
18
21
14
Explanation
$$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$$
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Point B lies on $$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$$
$$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$$
$$-3 \lambda-6=0$$
$$\lambda=-2$$
$$\mathrm{B} \Rightarrow(3,4,-1)$$
$$\mathrm{AB}=\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}$$
$$=\sqrt{16+36+144}$$
$$=\sqrt{196}=14$$
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