The distance of the point (1,0,2) from the point of intersection of the line x-23=y+14=z-2…

Question

The distance of the point $(1,0,2)$ from the point of intersection of the line $\dfrac{x-2}3=\dfrac{y+1}4=\dfrac{z-2}{12}$ and the plane $x-y+z=16$, is

Accepted Answer

General point on given line $$ \equiv P\left( {3r + 2,4r - 1,12r + 2} \right)$$

Point $P$ must satisfy equation of plane

$$\left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16$$

$11r + 5 = 16$

$r=1$

$$P\left( {3 \times 1 + 2,4 \times 1 - 1,12 \times 1 + 2} \right) = P\left( {5,3,14} \right)$$

Distance between $P$ and $(1,0,2)$

$$D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}}  = 13$$

Solutions · 1