Question

Find the distance of the point $P(3,4,4,)$ from the point, where the line joining the points $A(3,-4,-5)$ and $B(2,-3,1)$ intersected the plane $2x+y+z=7$

Answer 1

$P(3,4,4),A(3,-4,5),B(2,-3,1)$

Plane :$2x+y+z=7$

Line joining $A$ & $B$,

$\frac{x-3}{3-2}=\frac{y+4}{-4+3}=\frac{z+5}{-5-1}\Rightarrow \frac{x-3}{1}=\frac{y+4}{-1}=\frac{z+5}{-6}$

Let $(r+3,-r-4,-6r-5)$ be the point on line that intersects with plane .

$\therefore$ It must satisfy it $\Rightarrow 2(r+3)+(-r-4)+(-6r-5)=7$

$\Rightarrow r=-2$

$\therefore$ Point of intersection $=(1,-2,7)$

$\therefore$ distance from $P=\sqrt{{(3-1)}^2+{(4+2)}^2+{(4-7)}^2}=7$units.