Question

If the distance between a point $P$ and the point $(1,1,1)$ on the line $\frac{x-1}{3}=\frac{y-1}{4}=\frac{z-1}{12}$ is $13$, then the coordinates of $P$ are

• A. $(3,4,12)$
• B. $(\frac{3}{13},\frac{4}{13},\frac{12}{13})$
• C. $(4,5,13)$
• D. $(40,53,157)$

Answer 1

The correct option is C $(4,5,13)$

Let the given points be $A(1,1,1)$

Consider,

$\frac{x-1}{3}=\frac{y-1}{4}=\frac{z-1}{12}=\lambda$

$\begin{array}{c}x=3\lambda +1\\ \\ y=4\lambda +1\\ \\ z=12\lambda +1\end{array}$

General point on the line is

$3\lambda +1,4\lambda +1,12\lambda +1$

Given that,

$AP=13$

$\sqrt{{\left(3\lambda +1-1\right)}^2+{\left(4\lambda +1-1\right)}^2+{\left(12\lambda +1-1\right)}^2}=13$

$13\lambda =13$

$\lambda =1$

$\begin{array}{c}3\lambda +1=4\\ \\ 4\lambda +1=5\\ \\ 12\lambda +1=13\end{array}$

Therefore, required point $P$ is $(4,5,13)$.

Hence the correct option is $C$.