Question

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

Answer 1

$\begin{array}{c}Lines\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=k\\ LetQ=\left(5k-2,2k-1,2k+3\right)suchthat\\ PQ=3\sqrt{2}\\ fromdis\tan ceformula\\ {\left[\left(3k-2\right)-1\right]}^2+{\left[\left(2k-1\right)-2\right]}^2+{\left[\left(2k+3\right)-3\right]}^2=3\sqrt{2}\\ \Rightarrow {\left(3k-3\right)}^2+{\left(2k-3\right)}^2+{\left(2k\right)}^2={\left(3\sqrt{2}\right)}^2\\ \Rightarrow 9kh2-18k+9+4k^2-12k+9+4k^2=18\\ \Rightarrow 17k^2-30k=0\\ \Rightarrow k\left(17k-30\right)=0\\ \therefore k=0,\frac{30}{17}\\ ifk=0\\ Q=\left(3\times 0-2,2\times 0-1,2\times 0+3\right)\equiv \left(-2,-1,3\right)\\ Ifk=\frac{30}{17}\\ Q=\left(3\times \frac{30}{17}-2,2\times \frac{30}{17}-1,2\times \frac{30}{12}+3\right)\\ \equiv \left(\frac{56}{17},\frac{43}{17},\frac{11}{17}\right).\end{array}$