Mathematics · 2025
JEE Main · 28 January 2025, Shift 2 · Q16
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 i +4 j +7 k is:
The square of the distance of the point $\displaystyle \left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\displaystyle \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\displaystyle \hat{i}+4 \hat{j}+7 \hat{k}$ is :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 66$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.