Question

$\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are coplanar vectors such that $|\overrightarrow{a}|=2$, $|\overrightarrow{b}|=3$ & $\overrightarrow{a} \land \overrightarrow{b} = 30^{\circ}$. A unit vector $\overrightarrow{d}$ is perpendicular to all of them. If $(\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{c} \times \overrightarrow{d}) = \frac{\hat{i}}{6} - \frac{\hat{j}}{3} + \frac{\hat{k}}{3}$, then $|\overrightarrow{c}.\hat{i}| + |\overrightarrow{c}.\hat{j}| + |\overrightarrow{c}.\hat{k}|$ is equal to ____

• A. $\frac{5}{18}$
• B. $\frac{7}{18}$
• C. $\frac{11}{18}$
• D. $\frac{13}{18}$

Answer 1

Answer: (A)

$\frac{5}{18}$

Answer 2

The coplanarity of $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ implies $(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c} = 0$. The magnitude of the cross product $|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}|\sin(30^\circ) = 2 \times 3 \times \frac{1}{2} = 3$. Since $\overrightarrow{d}$ is a unit vector perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$, $\overrightarrow{a} \times \overrightarrow{b}$ is parallel to $\overrightarrow{d}$, so $\overrightarrow{a} \times \overrightarrow{b} = \pm 3\overrightarrow{d}$.

Using the vector triple product identity $\vec{X} \times (\vec{Y} \times \vec{Z}) = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{X} \cdot \vec{Y})\vec{Z}$, we simplify the given equation: $(\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{c} \times \overrightarrow{d}) = (\pm 3\overrightarrow{d}) \times (\overrightarrow{c} \times \overrightarrow{d})$ $= \pm 3 [\overrightarrow{d} \times (\overrightarrow{c} \times \overrightarrow{d})]$ $= \pm 3 [(\overrightarrow{d} \cdot \overrightarrow{d})\overrightarrow{c} - (\overrightarrow{d} \cdot \overrightarrow{c})\overrightarrow{d}]$ Since $\overrightarrow{d}$ is a unit vector, $\overrightarrow{d} \cdot \overrightarrow{d} = 1$. Since $\overrightarrow{d}$ is perpendicular to $\overrightarrow{c}$, $\overrightarrow{d} \cdot \overrightarrow{c} = 0$. So, the expression becomes $\pm 3 [1 \cdot \overrightarrow{c} - 0 \cdot \overrightarrow{d}] = \pm 3\overrightarrow{c}$.

Equating this to the given vector: $\pm 3\overrightarrow{c} = \frac{\hat{i}}{6} - \frac{\hat{j}}{3} + \frac{\hat{k}}{3}$ $\overrightarrow{c} = \pm \frac{1}{3} \left( \frac{\hat{i}}{6} - \frac{\hat{j}}{3} + \frac{\hat{k}}{3} \right)$

Let $\overrightarrow{c} = c_x \hat{i} + c_y \hat{j} + c_z \hat{k}$. We need to find $|c_x| + |c_y| + |c_z|$. In either case ($\pm$): $\overrightarrow{c} = \frac{1}{3} \left( \frac{\hat{i}}{6} - \frac{\hat{j}}{3} + \frac{\hat{k}}{3} \right)$ or $\overrightarrow{c} = -\frac{1}{3} \left( \frac{\hat{i}}{6} - \frac{\hat{j}}{3} + \frac{\hat{k}}{3} \right)$. The components of $\overrightarrow{c}$ will be $\pm \frac{1}{18}$, $\mp \frac{1}{9}$, $\pm \frac{1}{9}$. The absolute values of the components are $\frac{1}{18}$, $\frac{1}{9}$, $\frac{1}{9}$.

Therefore, $|\overrightarrow{c}.\hat{i}| + |\overrightarrow{c}.\hat{j}| + |\overrightarrow{c}.\hat{k}| = \left|\pm \frac{1}{18}\right| + \left|\mp \frac{1}{9}\right| + \left|\pm \frac{1}{9}\right| = \frac{1}{18} + \frac{1}{9} + \frac{1}{9} = \frac{1}{18} + \frac{2}{18} + \frac{2}{18} = \frac{5}{18}$.