Question

A point Q at a distance $3$ from the point P$(1,1,1)$ lying on the line joining the point A$(0,-1,3)$ and P, has the coodinates

• A. $(2,3,-1)$
• B. $(4,7,-5)$
• C. $(0,-1,3)$
• D. $(-2,-5,7)$

Answer 1

Answer: (A), (C)

$(2,3,-1)$
$(0,-1,3)$

The directional ratios of line joining the points $(1,1,1)$ and $(0,-1,3)$ are $1,2,-2$

The equation of line joining points $P,A$ is $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{-2}$

Any point on the line $AP$ should be in the form of $(t+1,2t+1,1-2t)$ $($ where $t$ is constant $)$

Given that the distance between $Q$ and $P$ is $3$ and $Q$ lies on $AP$

$\Rightarrow \sqrt{t^2+4t^2+4t^2}=3$

$\Rightarrow t^2=1$

$\Rightarrow t=\pm 1$

Therefore $Q$ can be $(2,3,-1)$ and $(0,-1,3)$

Therefore the correct options are $A$ and $C$