Question

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{c} \cdot (\vec{a} - 2\vec{b})$ is equal to __________.

Hint:

The cross product of two vectors is perpendicular to both of them. Use this to determine $\vec{c} \cdot \vec{b}$.

Answer 1

Correct Answer: 3

Alternatively, we can find the vector $\vec{c}$ explicitly. We use the identity for the vector triple product: $\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b}$.
Using the formula $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$:
$$ (\vec{a} \cdot \vec{c})\vec{a} - |\vec{a}|^2 \vec{c} = \vec{a} \times \vec{b} $$
Given $\vec{a} = (1, 1, 1)$, we have $|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$.
Also, $\vec{a} \cdot \vec{c} = 3$.
Now calculate $\vec{a} \times \vec{b}$:
$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1-1) - \hat{j}(-1-0) + \hat{k}(1-0) = -2\hat{i} + \hat{j} + \hat{k} $$
Substitute these into the vector triple product equation:
$$ 3\vec{a} - 3\vec{c} = -2\hat{i} + \hat{j} + \hat{k} $$
$$ 3(\hat{i} + \hat{j} + \hat{k}) - 3\vec{c} = -2\hat{i} + \hat{j} + \hat{k} $$
$$ 3\hat{i} + 3\hat{j} + 3\hat{k} + 2\hat{i} - \hat{j} - \hat{k} = 3\vec{c} $$
$$ 3\vec{c} = 5\hat{i} + 2\hat{j} + 2\hat{k} \Rightarrow \vec{c} = \frac{1}{3}(5, 2, 2) $$
Finally, compute $\vec{c} \cdot (\vec{a} - 2\vec{b})$:
$\vec{a} - 2\vec{b} = (\hat{i} + \hat{j} + \hat{k}) - 2(\hat{j} - \hat{k}) = \hat{i} - \hat{j} + 3\hat{k} = (1, -1, 3)$.
$$ \vec{c} \cdot (\vec{a} - 2\vec{b}) = \frac{1}{3}(5, 2, 2) \cdot (1, -1, 3) = \frac{1}{3}(5(1) + 2(-1) + 2(3)) = \frac{1}{3}(5 - 2 + 6) = \frac{9}{3} = 3 $$