Question

The coordinates of a point on the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{\sqrt{3}}$ at a distance $1$ unit from the point $(-1,-1,-\sqrt{3})$ are

• A. $(-\frac{3}{2},-\frac{7}{4},-\frac{5}{4}\sqrt{3})$
• B. $\left(\frac{3}{2},\frac{11}{4},\frac{\sqrt{3}}{4}\right)$
• C. $\left(\frac{1}{2},\frac{5}{4},-\frac{\sqrt{3}}{4}\right)$
• D. none of these.

Answer 1

Answer: (A)

$(-\frac{3}{2},-\frac{7}{4},-\frac{5}{4}\sqrt{3})$

Let the point be

$x=2t+1$
$y=3t+2$
$z=\sqrt{3}t$

Hence

$P=(2t+1,3t+2,\sqrt{3}t)$

$Q=(-1,-1,-\sqrt{3})$

Hence

$PQ=1$

$P{Q}^2=1$

$(2t+2{)}^2+(3t+3{)}^2+(\sqrt{3}t+\sqrt{3}{)}^2=1$

$(t+1{)}^2(4+9+3)=1$

$(t+1{)}^2(16)=1$

$(4t+4{)}^2=1$
$4t+4=\pm 1$
$4t+4=-1$

$\therefore t=\frac{-5}{4}$

Therefore

$P=(\frac{-3}{2},\frac{-7}{4},\frac{-5\sqrt{3}}{4})$