First slide

Scalar triple product of vectors

Question

If $\vec{a} and \vec{b}$ are non-zero non-collinear vectors, then $[\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} + [\vec{a} \vec{b} \hat{k}] \hat{k}$ is equal to

Moderate

A

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

$\vec{a} \times \vec{b}$
C

$\vec{a} - \vec{b}$
D

$\vec{b} - \vec{a}$

Solution

$\begin{array}{l} Let \vec{a} \times \vec{b} = x \hat{i} + y \hat{j} + z \hat{k} . Therefore, \\ [\vec{a} \vec{b} \hat{i}] = (\vec{a} \times \vec{b}) \cdot \hat{i} = x \\ [\vec{a} \vec{b} \hat{j}] = (\vec{a} \times \vec{b}) \cdot \hat{j} = y \\ [\vec{a} \vec{b} \vec{k}] = (\vec{a} \times \vec{b}) \cdot \hat{k} = z \\ Hence, [\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} \\ = [\vec{a} \vec{b} \hat{k}] \hat{k} = x \hat{i} + y \hat{j} + z \hat{k} = \vec{a} \times \vec{b} \end{array}$
$\begin{array}{l} hence [\vec{a} \vec{b} \hat{i}] \hat{i} + [\vec{a} \vec{b} \hat{j}] \hat{j} \\ = [\vec{a} \vec{b} \hat{k}] \hat{k} = x \hat{i} + y \hat{j} + z \hat{k} = \vec{a} \times \vec{b} \end{array}$

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