Question

If $\overline{a}$ is a vector of magnitude '2' and it is perpendicular to a unit vector $\overline{b}$ then $|\overline{a}\times(\overline{a}\times(\overline{a}\times(\overline{a}\times\overline{b})))|$ is equal to

• A. 2
• B. 4
• C. 8
• D. 16

Answer 1

Answer: (D)

16

Answer 2

Let $\overline{V} = \overline{a}\times(\overline{a}\times(\overline{a}\times(\overline{a}\times\overline{b}))))$. We are given $|\overline{a}| = 2$, $|\overline{b}| = 1$, and $\overline{a} \cdot \overline{b} = 0$.

We can evaluate this iteratively:

1. Let $\overline{v}_1 = \overline{a} \times \overline{b}$. Since $\overline{a}$ and $\overline{b}$ are perpendicular, $|\overline{v}_1| = |\overline{a}||\overline{b}|\sin(90^\circ) = (2)(1)(1) = 2$.
2. Let $\overline{v}_2 = \overline{a} \times \overline{v}_1$. Since $\overline{a}$ is perpendicular to $\overline{b}$, $\overline{a}$ is also perpendicular to $\overline{v}_1 = \overline{a} \times \overline{b}$. Thus, $|\overline{v}_2| = |\overline{a}||\overline{v}_1|\sin(90^\circ) = (2)(2)(1) = 4$.
3. Let $\overline{v}_3 = \overline{a} \times \overline{v}_2$. Since $\overline{a}$ is perpendicular to $\overline{v}_2$, $|\overline{v}_3| = |\overline{a}||\overline{v}_2|\sin(90^\circ) = (2)(4)(1) = 8$.
4. The final expression is $|\overline{a} \times \overline{v}_3|$. Since $\overline{a}$ is perpendicular to $\overline{v}_3$, $|\overline{a} \times \overline{v}_3| = |\overline{a}||\overline{v}_3|\sin(90^\circ) = (2)(8)(1) = 16$.

Alternatively, using the vector triple product formula $\overline{A} \times (\overline{B} \times \overline{C}) = (\overline{A} \cdot \overline{C})\overline{B} - (\overline{A} \cdot \overline{B})\overline{C}$: Let $\overline{X} = \overline{a} \times \overline{b}$. Since $\overline{a} \cdot \overline{b} = 0$, $\overline{X}$ is perpendicular to $\overline{a}$. $|\overline{X}| = 2$. $\overline{a} \times \overline{X} = \overline{a} \times (\overline{a} \times \overline{b}) = (\overline{a} \cdot \overline{b})\overline{a} - (\overline{a} \cdot \overline{a})\overline{b} = (0)\overline{a} - |\overline{a}|^2 \overline{b} = -4\overline{b}$. Let $\overline{Y} = -4\overline{b}$. $|\overline{Y}| = |-4||\overline{b}| = 4(1) = 4$. Note $\overline{Y}$ is perpendicular to $\overline{a}$ since $\overline{a} \cdot (-4\overline{b}) = -4(\overline{a} \cdot \overline{b}) = 0$. Next, $\overline{a} \times \overline{Y} = \overline{a} \times (-4\overline{b}) = -4(\overline{a} \times \overline{b}) = -4\overline{X}$. Let $\overline{Z} = -4\overline{X}$. $|\overline{Z}| = |-4||\overline{X}| = 4(2) = 8$. Note $\overline{Z}$ is perpendicular to $\overline{a}$ since $\overline{a} \cdot (-4\overline{X}) = -4(\overline{a} \cdot \overline{X}) = 0$. Finally, we need $|\overline{a} \times \overline{Z}|$. Since $\overline{a}$ is perpendicular to $\overline{Z}$, $|\overline{a} \times \overline{Z}| = |\overline{a}| |\overline{Z}| \sin(90^\circ) = (2)(8)(1) = 16$.