Finding the Perpendicular Vector to a Plane

To find a vector perpendicular to the plane containing points P(2,1,5), Q(−1,3,4), and R(3,0,6), we first determine two vectors lying within the plane. We can use vectors $\vec{PQ}$ and $\vec{PR}$.

Step 1: Define Vectors in the Plane

• Calculate $\vec{PQ}$: $\vec{PQ} = Q - P = (-1 - 2, 3 - 1, 4 - 5) = (-3, 2, -1)$
• Calculate $\vec{PR}$: $\vec{PR} = R - P = (3 - 2, 0 - 1, 6 - 5) = (1, -1, 1)$

Step 2: Calculate the Cross Product

A vector perpendicular to the plane is found by taking the cross product of $\vec{PQ}$ and $\vec{PR}$.

$ \vec{N} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 2 & -1 \\ 1 & -1 & 1 \end{vmatrix} $

Expanding the determinant:

$ \vec{N} = \mathbf{i} \left( (2)(1) - (-1)(-1) \right) - \mathbf{j} \left( (-3)(1) - (-1)(1) \right) + \mathbf{k} \left( (-3)(-1) - (2)(1) \right) $

$ \vec{N} = \mathbf{i} (2 - 1) - \mathbf{j} (-3 + 1) + \mathbf{k} (3 - 2) $

$ \vec{N} = \mathbf{i}(1) - \mathbf{j}(-2) + \mathbf{k}(1) $

$ \vec{N} = 1\mathbf{i} + 2\mathbf{j} + 1\mathbf{k} $

Therefore, the vector perpendicular to the plane is $\mathbf{i} + 2\mathbf{j} + \mathbf{k}$.

Step 3: Match with Options

The calculated vector $\mathbf{i} + 2\mathbf{j} + \mathbf{k}$ corresponds to Option D.