The distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to the line x2…

Question

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\dfrac x2=\dfrac y3=\dfrac z{-6}$ is:

Accepted Answer

Equation of line parallel to $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ passes through $(1, - 2,3)$ is

$${{x - 1} \over 2} = {{y + 2} \over 3} = {{z - 3} \over { - 6}} = r$$

$x = 2r + 1$

$y = 3r - 2$,

$z =  - 6r + 3$

A point on whole line = (2r + 1, 3r -- 2, -- 6r + 3).

This point lies on plane x -- y + 2 = 5

so, $2r + 1 - 3r + 2 - 6r + 3 = 5$

$$ \Rightarrow $$ $$r = {1 \over 7}$$

$$ \therefore $$ $$x = {9 \over 7}$$, $$y = {{ - 11} \over 7}$$, $$z = {{15} \over 7}$$

Distance is = $$\sqrt {{{\left( {{9 \over 7} - 1} \right)}^2} + {{\left( {2 - {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}} $$

$$ = \sqrt {{{\left( {{2 \over 7}} \right)}^2} + {{\left( {{3 \over 7}} \right)}^2} + {{\left( {{6 \over 7}} \right)}^2}} $$

$$ = {1 \over 7}\sqrt {4 + 9 + 36} $$ = 1

Solutions · 1