Question

The co-ordinates of point, where the line $\overline{r}=(\hat{i}+2\hat{j}-3\hat{k})+\lambda (2\hat{i}-3\hat{j}+4\hat{k})$ meets the plane $2x+4y-z=1$ are ?

• A. $(3,-1,1)$
• B. $(3,1,9)$
• C. $(1,2,9)$
• D. $(5,1,8)$

Answer 1

Answer: (A)

$(3,-1,1)$

Given,

Line, $\overline{r}=(\hat{i}+2\hat{j}-3\hat{k})+\lambda (2\hat{i}-3\hat{j}+4\hat{k})$

Plane, $2x+4y-z=1$

The value of $\lambda$ can be solved by replacing the values of x,y and z in the planar equation with the linear equations,

$2(1+2\lambda )+4(2-3\lambda )+(-3+4\lambda )-1=0$

$2+4\lambda +6-12\lambda -3+4\lambda -1=0$

$4-4\lambda =0$

$\lambda =1$

The point where the plane and the line intersect can be computed by using $\lambda =1$

$x=1+2\left(1\right)=3$

$y=2-3\left(1\right)=-1$

$z=-3+4\left(1\right)=1$

The point is $(3,-1,1)$