The co-ordinates of the point, where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ meets the plane $2 x+4 y-z=3$, are

Question

Aicharya · Accepted Answer

Given line is $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ Let $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}=\lambda$ $\begin{array}{ll} 	herefore \quad & x-1=2 \lambda, y-2=-3 \lambda, \mathrm{z}+5=4 \lambda \ & x=2 \lambda+1, y=-3 \lambda+2, \mathrm{z}=4 \lambda-5 \ 	herefore \quad & 2 x+4 y-\mathrm{z}=3 \ & \Rightarrow 2(2 \lambda+1)+4(-3 \lambda+2)-(4 \lambda-5)=3 \ & \Rightarrow 4 \lambda+2-12 \lambda+8-4 \lambda+5=3 \ & \Rightarrow-12 \lambda=3-15 \ & \Rightarrow \lambda=1 \ 	herefore \quad & x=3, y=-1, \mathrm{z}=-1, \end{array}$ $	herefore \quad$ Required co-ordinates are: $(3,-1,-1)$

The co-ordinates of the point, where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ meets the plane…

Solution