Question

The line passing through the points $(1,2,-1)$ and $(3,-1,2)$ meets the $yz$plane at which one of the following points-

• A. $\left(0,-\frac{7}{2},\frac{5}{2}\right)$
• B. $\left(0,\frac{7}{2},\frac{5}{2}\right)$
• C. $\left(0,-\frac{7}{2},-\frac{5}{2}\right)$
• D. $\left(0,\frac{7}{2},-\frac{5}{2}\right)$

Answer 1

The correct option is D $\left(0,\frac{7}{2},-\frac{5}{2}\right)$

$\begin{array}{c}{l}_1=3-1\\ m_1=-1-2\\ n_1=2+1\\ l_2=3-0\\ m_2=-1-y\\ n_2=2-z\\ \frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2}\\ \frac{2}{3}=\frac{-3}{-1-y}=\frac{3}{2-z}\\ \therefore \frac{2}{3}=\frac{3}{1+y}\\ 2+2y=9\Rightarrow y=\frac{7}{2}\\ and\\ \frac{2}{3}=\frac{3}{2-z}\\ 1-2z=9\Rightarrow z=\frac{-5}{2}\\ \therefore points\equiv \left(0,\frac{7}{2},-\frac{5}{2}\right)\end{array}$