# 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)

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Solution

First, We will find vector form of (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

$$\vec{u_{1}}$$ = (1, 4, 6) – (-2, 5, -1) = (3, -1, 7)

$$\vec{u_{2}}$$ = (1, 4, 6) – (1, -1, 1) = (0, 5, 5)

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 $$\vec{u_{1}}$$ and $$\vec{u_{2}}$$ to find a vector $$\vec{u}$$ perpendicular to the plane containing them.

$$\Rightarrow \vec{u} = \vec{u_{2}} \times \vec{u_{2}}$$

$$\Rightarrow \vec{u} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 3 & -1 & 7\\ 0 & 5 & 5 \end{vmatrix}$$

$$\Rightarrow 40\vec{i} – 15\vec{j} + 15\vec{k}$$