Question

if the line X+1/2=y-m/3=z-4/6 lies in the plane 3x-14y+6z+49=0, then the value of m is​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{Line\;\dfrac{x+1}{2}=\dfrac{y-m}{3}=\dfrac{z-4}{6}\;lies\;in\;the\;plane\;3x-14y+6z+49=0}$

$\underline{\textbf{To find:}}$

$\textsf{The value of m}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{x+1}{2}=\dfrac{y-m}{3}=\dfrac{z-4}{6}}$

$\textsf{Clearly (-1,m,4) lies on this line}$

$\textsf{Since the line lies in the plane,}$

$\textsf{every point of the line lying in the plane}$

$\implies\mathsf{(-1,m,4)\;lies\;in\;the\;plane\;3x-14y+6z+49=0}$

$\implies\mathsf{3(-1)-14(m)+6(4)+49=0}$

$\implies\mathsf{-3-14m+24+49=0}$

$\implies\mathsf{-14m+70=0}$

$\implies\mathsf{-14m=-70}$

$\implies\mathsf{m=\dfrac{-70}{-14}}$

$\implies\boxed{\mathsf{m=5}}$

Answer 2

Answer:

since the line lies on the plane it satisfies the equation of plane