Question

Find the vector equation of plane which is at a distance of 5 units from the origin and its normal vector is 2i^ -3j^ +6k^.​

Answer 1

vector equation of a plane is of the form

r.n^ = d

here n = 2i^-3j^+6k^

so,

$|n | = \sqrt{ {2}^{2} + { (- 3)}^{2} + {6}^{2} }$

$|n | = \sqrt{4 + 9 + 36}$

= 7

so the equation will be

$r.( \frac{2}{7} i - \frac{3}{7} j + \frac{6}{7} k) = 5$