Vectors — practice questions

(a) We set up the determinant to calculate the cross product: $\mathbf{p} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 4 & -2 \end{vmatrix}$ $= \mathbf{i}((-1)(-2) - (3)(4)) - \mathbf{j}((2)(-2) - (3)(1)) + \mathbf{k}((2)(4) - (-1)(1))$ $= \mathbf{i}(2 - 12) - \mathbf{j}(-4 - 3) + \mathbf{k}(8 + 1)$ $= -10\mathbf{i} + 7\mathbf{j} + 9\mathbf{k}$

(b) The area of the parallelogram is the magnitude of the vector product. Area $= |\mathbf{p} \times \mathbf{q}| = |-10\mathbf{i} + 7\mathbf{j} + 9\mathbf{k}|$ $= \sqrt{(-10)^2 + 7^2 + 9^2}$ $= \sqrt{100 + 49 + 81}$ $= \sqrt{230}$ The exact area is $\sqrt{230}$ square units.

First, we find three vectors originating from a common vertex, say A. $\vec{AB} = \mathbf{b} - \mathbf{a} = (3-1)\mathbf{i} + (3-2)\mathbf{j} + (0-1)\mathbf{k} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$ $\vec{AC} = \mathbf{c} - \mathbf{a} = (2-1)\mathbf{i} + (6-2)\mathbf{j} + (4-1)\mathbf{k} = \mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$ $\vec{AD} = \mathbf{d} - \mathbf{a} = (5-1)\mathbf{i} + (5-2)\mathbf{j} + (5-1)\mathbf{k} = 4\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$

The volume of the tetrahedron is $\frac{1}{6}|\vec{AB} \cdot (\vec{AC} \times \vec{AD})|$. We calculate the scalar triple product using a determinant: $\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = \begin{vmatrix} 2 & 1 & -1 \\ 1 & 4 & 3 \\ 4 & 3 & 4 \end{vmatrix}$ $= 2((4)(4) - (3)(3)) - 1((1)(4) - (3)(4)) + (-1)((1)(3) - (4)(4))$ $= 2(16 - 9) - 1(4 - 12) - 1(3 - 16)$ $= 2(7) - 1(-8) - 1(-13)$ $= 14 + 8 + 13 = 35$

Volume $= \frac{1}{6} |35| = \frac{35}{6}$ cubic units.