Question

A point on the line $\frac{x+2}{1}=\frac{y-3}{-4}=\frac{z-1}{2\sqrt{2}}$ at a distance 6 from the point (2, 3, 1) is

• A. $(4-21,1+12\sqrt{2})$
• B. $\left(\frac{-4}{5},\frac{-9}{5},1\right)$
• C. $\left(\frac{-16}{5},\frac{39}{5},\frac{5-12\sqrt{2}}{5}\right)$
• D. $\left(\frac{-16}{5},-21,1+12\sqrt{2}\right)$

Answer 1

Answer: (B)

$\left(\frac{-4}{5},\frac{-9}{5},1\right)$

$(\lambda -2,-4\lambda +3,2\sqrt{2}\lambda +1)\leftrightarrow (-2,3,1)$
$(\lambda -2+2{)}^2+(-4\lambda +3-3{)}^2+2\sqrt{2}\lambda +1-1{)}^2=36$
${\lambda }^2+16{\lambda }^2+8{\lambda }^2=36$
$\lambda =\pm \frac{6}{5}$
$\therefore point=(\frac{-4}{5},\frac{-9}{5},1)$