Q85 / 180 · KCET 2020 · Mathematics
The distance of the point $(1,2,-4)$ from the line $\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}$ is
- A$\frac{293}{7}$
- Correct answerB$\frac{\sqrt{293}}{7}$
- C$\frac{293}{49}$
- D$\frac{\sqrt{293}}{49}$
Solution
Given, point $(1,2,-4)$.
$\begin{aligned}
&\text { and line } \frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}=\lambda \text { (let) } \\
&\begin{aligned}
& \therefore & x &=2 \lambda+3 \\
& \therefore & x &=2 \lambda+3, y=3 \lambda+3, z=6 \lambda-5
\end{aligned}
\end{aligned}$
Let $P(2 \lambda+3,3 \lambda+3,6 \lambda-5)$ be the foot of perpendicular from $A(1,2,4)$
$\therefore$ Direction ratios of $A P$ are
$2 \lambda+3-1,3 \lambda+3-2,6 \lambda-5+4$
i.e., $2 \lambda+2,3 \lambda+1,6 \lambda-1$
And direction ratios of given line are $2,3,6$.
Since, $\Lambda P \perp$ given line
$\begin{array}{cc}\therefore & 2(2 \lambda+2)+3(3 \lambda+1)+6(6 \lambda-1)=0 \\ \Rightarrow & 4 \lambda+4+9 \lambda+3+36 \lambda-6=0\end{array}$
$\Rightarrow \quad 49 \lambda+1=0$
$\Rightarrow \quad \lambda=-\frac{1}{49}$
$\therefore P(2 \lambda+3,3 \lambda+3,6 \lambda-5)$
$\therefore$ Distance of the point $(1,2,-4)$ from the given line,
$\begin{aligned}
A P^{2} &=(2 \lambda+2)^{2}+(3 \lambda+1)^{2}+(6 \lambda-1)^{2} \\
&=4 \lambda^{2}+4+8 \lambda+9 \lambda^{2}+1+6 \lambda+36 \lambda^{2}+1-12 \lambda \\
&=49 \lambda^{2}+6+2 \lambda \\
&=49\left(-\frac{1}{49}\right)^{2}+6+2\left(-\frac{1}{49}\right)
\end{aligned}$
$\begin{aligned} &=\frac{1}{49}+6-\frac{2}{49} \\ &=6-\frac{1}{49}=\frac{293}{49} \\ \Rightarrow A P &=\frac{\sqrt{293}}{7} \end{aligned}$
Hence, the correct answer is option: 'B'.