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 point plane 4x + 12y - 3z + 1 = 0.

Answer 1

Consider the line

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

Let,

$\frac{x+2}{3}=\frac{y+3/2}{2}=\frac{z+4/3}{5/3}=t$ --- $\left(i\right)$

Let the line $\left(i\right)$ passes through,

$x=3t-2;y=2t-\frac{3}{2};z=\frac{5}{3}t-\frac{4}{3}$

and parallel to $4x+12y-3z+1=0$

$4\left(3t-2\right)+12\left(2t-\frac{3}{2}\right)-3\left(\frac{5}{3}t-\frac{4}{3}\right)+1=0.$

Therefore,

$t=\frac{21}{33}$

therefore points is $(-\frac{1}{11},-\frac{15}{66},-\frac{27}{99})$

Therefore distance between $\left(-2,3,-4\right)$ and $\left(-\frac{1}{11},-\frac{15}{66},-\frac{27}{99}\right)$

$=\sqrt{{\left(-\frac{1}{11}+2\right)}^2+{\left(-\frac{15}{66}-3\right)}^2+{\left(-\frac{27}{99}+4\right)}^2}$

$=\sqrt{{\left(\frac{21}{11}\right)}^2+{\left(-\frac{183}{66}\right)}^2+{\left(\frac{369}{99}\right)}^2}$

$=\sqrt{25}$

$=5$ sq. units