Question

4.The distance of the point (1,-2,8) from the plane 2x-3y+6z=63 is​

Answer 1

ANSWER :–

Distance = 1 unit

EXPLANATION :–

GIVEN :–

• A point (1,-2,8) and a plane 2x - 3y + 6z = 63

TO FIND :–

Distance between point and plane.

SOLUTION :–

• The distance from a point (a, b,c) to the plane Ax + By + Cz + D = 0 is –

$\\ { \bold{ \implies{ \:Distance \: \: = | \frac{A(a) +B(b) +C(c) +D}{ \sqrt{ {A}^{2} + {B}^{2} + {C}^{2} } } | }}}$

• Now put the values –

$\\ \implies \: Distance = | \frac{2(1) - 3( - 2) + 6(8) - 63}{ \sqrt{ {(2)}^{2} + {( - 3)}^{2} + {(6)}^{2} } }| \\ \\ \implies \: Distance = | \frac{2 + 6 + 48 - 63}{ \sqrt{4 + 9 + 36} } | \\ \\ \implies \: Distance = | \frac{ - 7}{ \sqrt{49} } | \\ \\ \implies \: Distance = 1 \: \: unit$