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:
Step-by-Step Solution
Key Concept: Write the line through $P=\left(\tfrac{15}{7},\tfrac{32}{7},7\right)$ in direction $(1,4,7)$, find its intersection $Q$ with the given line, then compute $|PQ|^2$.
Line from $P$: $\dfrac{x-\tfrac{15}{7}}{1}=\dfrac{y-\tfrac{32}{7}}{4}=\dfrac{z-7}{7}=\lambda$.
General point: $\left(\lambda+\tfrac{15}{7},4\lambda+\tfrac{32}{7},7\lambda+7\right)$.
For $Q$ on given line: $\dfrac{(\lambda+\tfrac{15}{7})+1}{3}=\dfrac{7\lambda+7+5}{7}$ gives $\lambda=-1$.
$Q=\left(\tfrac{8}{7},\tfrac{4}{7},0\right)$.
$|PQ|^2=\left(\tfrac{15}{7}-\tfrac{8}{7}\right)^2+\left(\tfrac{32}{7}-\tfrac{4}{7}\right)^2+(7-0)^2=1+16+49=66$.
Correct Answer: 4