Find a vector perpendicular to the plane that passes through the points P(1, 4, 6), Q(-2, 5, -1) and R(1, -1, 1).

Concept:

To find a vector perpendicular to the find two planes, first, find the vector in the plane and then take their cross product.

Solution:

Given, the plane is passing through the points P(1, 4, 6), Q(-2, 5, -1) and R(1, -1, 1).

Step (1) Find two vectors in the plane.

We will do this by finding the vector from (1, 4, 6) (-2, 5, -1) and (1, 4, 6) (1, -1, 1).

As all three points are in the plane, so will each of those vectors.

\(\overrightarrow{u_1}=(1,4,6)-(-2,5,-1)=(3,-1,7)\)

\(\overrightarrow{u_2}=(1,4,6)-(1,-1,1)=(0,5,5)\)

Step (2) Find a vector perpendicular to the plane.

If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross-product of two vectors produces a vector perpendicular to both, we will use the cross-product of \(\overrightarrow{u_1}\ and \ \overrightarrow{u_2}\)  to find a vector \(\overrightarrow{u}\) perpendicular to the plane containing them.

\(\Rightarrow \overrightarrow{u}=\overrightarrow{u_1} \times \overrightarrow{u_2}\)

\(\Rightarrow \overrightarrow{u}=\begin{vmatrix} i &j & k \\ 3& -1 & 7 \\ 0&5 & 5\\ \end{vmatrix}\)

\(\Rightarrow \overrightarrow{u}=i(-5-35)-j(15-0)+k(15-0)\)

\(\Rightarrow \overrightarrow{u}=-40i-15j+15k\)

Hence, the vector perpendicular to the plane that passes through the points P(1, 4, 6), Q(-2, 5, -1), and R(1, -1, 1) is (-40, -15, 15).