Question

The point of intersection of the line segment joining a(2,3,4) and b(-3,5,-4)and the yz plane is

Answer 1

$\textbf{Given:}$

$\textsf{Points are A(2,3,4) and B(-3,5,-4)}$

$\textbf{To find:}$

$\textsf{Point of intersection of line segment joining A and B with yz plane}$

$\textbf{Solution:}$

$\textsf{The equation of line joining A and B is}$

$\mathsf{\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}=\dfrac{z-z_1}{z_2-z_1}}$

$\mathsf{\dfrac{x-2}{-3-2}=\dfrac{y-3}{5-3}=\dfrac{z-4}{-4-4}}$

$\mathsf{\dfrac{x-2}{-5}=\dfrac{y-3}{2}=\dfrac{z-4}{-8}}$

$\textsf{It meets yz plane}$

$\mathsf{That\;is\;x=0}$

$\mathsf{\dfrac{-2}{-5}=\dfrac{y-3}{2}=\dfrac{z-4}{-8}}$

$\mathsf{\dfrac{2}{5}=\dfrac{y-3}{2}=\dfrac{z-4}{-8}}$

$\implies\mathsf{\dfrac{2}{5}=\dfrac{y-3}{2}}$

$\implies\mathsf{\dfrac{4}{5}=y-3}$

$\implies\mathsf{y=\dfrac{4}{5}+3}$

$\implies\mathsf{y=\dfrac{19}{5}}$

$\mathsf{and}$

$\mathsf{\dfrac{2}{5}=\dfrac{z-4}{-8}}$

$\implies\mathsf{\dfrac{-16}{5}=z-4}$

$\implies\mathsf{z=\dfrac{-16}{5}+4}$

$\implies\mathsf{z=\dfrac{4}{5}}$

$\therefore\textsf{The point of intersection is}$

$\mathsf{\left(0,\dfrac{19}{5},\dfrac{4}{5}\right)}$