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 :

Correct answer: D. 66 aligned & L=(x+1)/(3)=(y+3)/(5)=(z+5)/(7) & P Q=(x-(15)/(7))/(1)=(y-(32)/(7))/(4)=(z-7)/(7)=λ aligned Q (λ+(15)/(7), 4 λ+(32)/(7), 7 λ+7) Since Q lies on line L aligned & So, (λ+(15)/(7)+1)/(3)=(7 λ+7+5)/(7) & 7 λ+22=21 λ+36 & λ=-1 & Point Q((8)/(7), (4)/(7), 0) & PQ= ((15)/(7)-(8)/(7))^2+((32)/(7)-(4)/(7))^2+(7-0) & PQ= 66 & (PQ)^2=66 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Three Dimensional Geometry

The square of the distance of the point ((15)/(7), (32)/(7), 7)...

The square of the distance of the point

\left(\frac{15}{7}, \frac{32}{7}, 7\right)

from the line

\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}

in the direction of the vector

\hat{i}+4 \hat{j}+7 \hat{k}

is :

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

Options

A $54$
B $44$
C $41$
D $66$

Correct answer

D. $66$

Step-by-step solution

\begin{aligned} & L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \\ & P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda\end{aligned}

\Rightarrow \mathrm{Q}

\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)

Since Q lies on line L

\begin{aligned} & \text{ So, } \frac{\lambda+\frac{15}{7}+1}{3}=\frac{7 \lambda+7+5}{7} \\ & \Rightarrow 7 \lambda+22=21 \lambda+36 \\ & \Rightarrow \lambda=-1 \\ & \therefore \text{ Point } Q\left(\frac{8}{7}, \frac{4}{7}, 0\right) \\ & \mathrm{PQ}=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^2+\left(\frac{32}{7}-\frac{4}{7}\right)^2+(7-0)} \\ & \mathrm{PQ}=\sqrt{66} \\ & \Rightarrow(\mathrm{PQ})^2=66 \end{aligned}

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