Vectors: Question 10

Worked solution

Part (a): Finding $k$

Two vectors are perpendicular exactly when their scalar product is zero:

$\mathbf{a}\cdot\mathbf{b} = (3)(2) + (k)(-1) + (-2)(4) = 6 - k - 8 = -k - 2$

Setting this equal to zero:

$-k - 2 = 0 \implies k = -2$

Check by recomputing independently: substituting $k=-2$ back into $\mathbf{a}\cdot\mathbf{b}$: $(3)(2) + (-2)(-1) + (-2)(4) = 6 + 2 - 8 = 0$ ✓. The scalar product is indeed zero, confirming $k=-2$.

Part (b): Finding $|\mathbf{a}|$

With $k=-2$, $\mathbf{a} = \begin{pmatrix}3\\-2\\-2\end{pmatrix}$, so:

$|\mathbf{a}| = \sqrt{3^2 + (-2)^2 + (-2)^2} = \sqrt{9+4+4} = \sqrt{17}$

Check by recomputing in a different order: $(-2)^2 + (-2)^2 + 3^2 = 4+4+9 = 17$, so $|\mathbf{a}| = \sqrt{17}$ again, the same result.

Part (c): Unit vector in the direction of $\mathbf{a}$

Dividing $\mathbf{a}$ by its own magnitude:

$\frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{\sqrt{17}}\begin{pmatrix}3\\-2\\-2\end{pmatrix} = \frac{3}{\sqrt{17}}\mathbf{i} - \frac{2}{\sqrt{17}}\mathbf{j} - \frac{2}{\sqrt{17}}\mathbf{k}$

Check: the magnitude of this vector must equal $1$:

$\left(\frac{3}{\sqrt{17}}\right)^2 + \left(\frac{-2}{\sqrt{17}}\right)^2 + \left(\frac{-2}{\sqrt{17}}\right)^2 = \frac{9+4+4}{17} = \frac{17}{17} = 1 \checkmark$

Final answers

• (a) $k = -2$
• (b) $|\mathbf{a}| = \sqrt{17}$
• (c) $\dfrac{3}{\sqrt{17}}\mathbf{i} - \dfrac{2}{\sqrt{17}}\mathbf{j} - \dfrac{2}{\sqrt{17}}\mathbf{k}$