Question

find a vector perpendicular to the plane that passes through P(1,-2,5), Q(-2,3,-1) and R(-3,1,2).

Answer 1

Given :

Coordinates of P = (1, -2 , 5 )

Coordinates of Q = (-2 , 3 , -1 )

Coordinates of R = (-3 , 1, 2 )

To Find :

A vector which is perpendicular to the plane that passes through these point P , Q and R  = ?

Solution :

The vector $\vec {PQ}$ = (-2-1)i + (3 + 2)j + (-1-5)k = -3i + 5j - 6k

And the vector $\vec {QR}$ = (-3+2)i + (1-3)j + (2 + 1)k = -i -2j +3k

Now the equation of normal to the plane passing through these three points is given as :

= $\vec {PQ}\times \vec {QR}$

So, $\vec PQ}\times \vec {QR} = \left|\begin{array}{ccc}i&j&k\\-3&5&-6\\-1&-2&3\end{array}\right|$

                      =i (15 -12 ) - j(-9-6) + k(6 +5)

                      = 3i  + 15j + 11k

Hence,, the vector perpendicular to the plane that passes  through the given three points P , Q and R is  3i + 15j + 11k .