The square of the distance of the point ((15)/(7),(32)/(7),7 ) from the line (x+1)/(3)=(y+3)/(5)=(z+5)/(7) in the direction of the vector hati+4hatj+7hatk is:

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JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1

The square of the distance of the point $\left(\dfrac{15}{7},\dfrac{32}{7},7\right)$ from the line $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ in the direction of the vector $\hat{i}+4\hat{j}+7\hat{k}$ is:

3D Geometry

$66$✓ Correct answer$54$$41$$44$

Solution

Moving from the point along $(1,4,7)$: $\left(\frac{15}7+t,\frac{32}7+4t,7+7t\right)$ meets the line at $t=-1$, i.e. $\left(\frac87,\frac47,0\right)$. The squared distance is $1+16+49=66$.

3D Geometry - JEE Mathematics question (JEE Main 2025 (Online) 28th January Evening Shift)

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