Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Since direction cosines of PQ are equal and positive. $\therefore$ The d.r.s. of PQ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$ $\therefore$ The equation of the line $\mathrm{PQ}$ is $$\begin{array}{rl}& \frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y+1}{\frac{1}{\sqrt{3}}}=\frac{\mathrm{z}-2}{\frac{1}{\sqrt{3}}}\\ & \Rightarrow x-2=y+1=\mathrm{z}-2=\mathrm{k},\text{ say }\end{array}$$ $\therefore$ Co-ordinates of the point $\mathrm{Q}$ are $$(\mathrm{k}+2,\mathrm{k}-1,\mathrm{k}+2)$$ The point $\mathrm{Q}$ lies on the plane $2x+y+z=9$ $$\begin{array}{ll}\therefore & 2(\mathrm{k}+2)+\mathrm{k}-1+\mathrm{k}+2=9\\ & \Rightarrow 4\mathrm{k}+5=9\Rightarrow \mathrm{k}=1\\ \therefore & \mathrm{Q}\equiv (3,0,3)\\ \therefore & \mathrm{PQ}=\sqrt{(3-2{)}^2+(0+1{)}^2+(3-2{)}^2}\\ & =\sqrt{1+1+1}=\sqrt{3}\end{array}$$