Question

The co-ordinates of a point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4\sqrt{14}$ from the point $(1, - 1, 0)$ are

• A. $(9, - 13, 4)$
• B. $(- 9, 13, 4)$
• C. $(9, 13,-4)$
• D. none of these

Answer 1

Answer: (A)

$(9, - 13, 4)$

Answer 2

Here $\frac{x-2}{2}=\frac{y+1}{3}=z$ ...(1)
is given straight line
Let P = (1, - 1, 0) on the straight line.
Here direction ratios of line (1) are (2, - 3,1)
$\therefore$ direction cosines of line (1) are
$\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
$\Rightarrow$ Equation of line (1) may be written as
$\frac{x-1}{2 - \sqrt{4}}=\frac{y-1}{-3 /\sqrt{14}}=\frac{z}{1/ \sqrt{14}}$ ....(2)

Co-ordinates of any point on line (2) may be taken as
$\left(\frac{2r}{\sqrt{14}}+1, \frac{-3r}{\sqrt{14} } -1, \frac{1r}{\sqrt{14}}\right)$
Let Q = $\left(\frac{2r}{\sqrt{14}}+1, \frac{-3r}{\sqrt{14}} - 1, \frac{1r}{\sqrt{14}}\right)$
Given $|\overrightarrow{r}| = 4 \sqrt{14}$ $\therefore \, r = \pm 4 \sqrt{14}$ $\therefore$
Putting the value of $r$, we get
Q = $[(9, - 13, 4)]$ and Q$(-7, 11-4)$