Question

A point on the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+1}{3}$ at a distance$\sqrt{6}$ from the origin is

• A. $\left(\frac{-5}{7},\frac{-10}{7},\frac{13}{7}\right)$
• B. $(1,2,-1)$
• C. $\left(\frac{5}{7},\frac{10}{7},\frac{-13}{7}\right)$
• D. $(-1,-2,1)$

Answer 1

Answer: (B), (C)

$(1,2,-1)$
$\left(\frac{5}{7},\frac{10}{7},\frac{-13}{7}\right)$

Given line is, $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+1}{3}=k$(assume)
So any point on this line is $P(k+1,2k+2,3k-1)$
Distance of any point on this line from the origin is,
$\sqrt{(k+1{)}^2+(2k+2{)}^2+(3k-1{)}^2}=\sqrt{6}$ (given)

$\Rightarrow 14k^2+4k=0$
$\Rightarrow k=0,\frac{-2}{7}$

Therefore, point on the line is,
$(1,2,-1)$ or $(\frac{5}{7},\frac{10}{7},-\frac{13}{7})$