First slide

Scalar triple product of vectors

Question

For any two vectors $\vec{a} and \vec{b}, (\vec{a} \times \hat{i}) \cdot (\vec{b} \times \hat{i}) + (\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j}) + (\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k})$ is always equal to

Moderate

A

$\vec{a} \cdot \vec{b}$
B

$2 \vec{a} \cdot \vec{b}$
C

zero
D

None of these

Solution

$\begin{aligned} (\vec{a} \times \hat{i}) \cdot (\vec{b} \times \hat{i}) = |\begin{array}{ll} \vec{a} \cdot \vec{b} & \vec{a} \cdot \hat{i} \\ \vec{b} \cdot \hat{i} & \hat{i} \cdot \hat{i} \end{array}| \\ = (\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \hat{i}) (\vec{b} \cdot \hat{i}) \end{aligned}$
$Similarly, (\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j}) = (\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \hat{j}) (\vec{b} \cdot \hat{j})$
$and (\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k}) = \vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{k}) (\vec{b} \cdot \hat{k})$
$Let \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$
$\begin{array}{l} ∴ (\vec{a} \cdot \hat{i}) = a_{1}, \vec{a} \cdot \hat{j} = a_{2}, \\ \vec{a} \cdot \hat{k} = a_{3}, \vec{b} \cdot \hat{i} = b_{1}, \vec{b} \cdot \hat{j} = b_{2}, (\vec{b} \cdot \hat{k}) = b_{3} \\ \Rightarrow (\vec{a} \times \hat{i}) \cdot (\vec{b} \times \hat{i}) + (\vec{a} \times \hat{j}) \cdot (\vec{b} \times \hat{j}) + (\vec{a} \times \hat{k}) \cdot (\vec{b} \times \hat{k}) \end{array}$
$\begin{array}{l} = 3 \vec{a} \cdot \vec{b} - (a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}) \\ = 3 \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b} = 2 \vec{a} \cdot \vec{b} \end{array}$

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