Question

Find the distance of the point (-2,3,-4) from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane 4x+12y-3z+1=0.

Answer 1

Consider the equation,

$\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}=\lambda$

Therefore, any point on this line is of the form,

$\left[3\lambda -2,\frac{4\lambda -3}{2},\frac{5\lambda -4}{3}\right]$

Now, the line from the point $(-2,3,-4)$ is $3\lambda ,\frac{\left(4\lambda -9\right)}{2},\frac{5\lambda +8}{3}$

The equation of a plane is $4x+12y-3z+1=0$

Thus, Direction ratio of normal is $4,12,-3$

Therefore,

$\begin{array}{c}4\cdot 3\lambda +12\left[\frac{\left(4\lambda -9\right)}{2}\right]-3\left[\left(\frac{5\lambda +8}{3}\right)\right]=0\\ 12\lambda +24\lambda -54-5\lambda -8=0\\ 31\lambda =62\\ \lambda =2\end{array}$

Hence, the required coordinates is $(4,5/2,2)$

Hence, the distance between coordinates $(4,5/2,2)$ and $(2,3,-4)$ is $17/2unit$