First slide

Vector triple product

Question

If $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}, \vec{b} = y \hat{i} + z \hat{j} + x \hat{k} and \vec{c} = z \hat{i} + x \hat{j} + y \hat{k}, then \vec{a} \times (\vec{b} \times \vec{c})$ is

Moderate

A

parallel to $(y - z) \hat{i} + (z - x) \hat{j} + (x - y) \hat{k}$
B

orthogonal to $\hat{i} + \hat{j} + \hat{k}$
C

orthogonal to $(y + z) \hat{i} + (z + x) \hat{j} + (x + y) \hat{k}$
D

orthogonal to $x \hat{i} + y \hat{j} + z \hat{k}$

Solution

$\begin{array}{l} \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \\ = (yz + yx + zx) {(y - z) \hat{i + (z - x) \hat{j} + (x - y) \hat{k}}} \end{array}$
Clearly this vector is parallel to $(y - z) \hat{i} + (z - x) \hat{j} + (x - y) \hat{k}$
It is orthogonal to $\hat{i} + \hat{j} + \hat{k} as (y - z) (1) + (z - x) (1)$ $+ (x - y) (1) = 0$
It is orthogonal to $(y + z) \hat{i} + (z + x) \hat{j} + (x + y) \hat{k}$
$\begin{matrix} as, (y - z) (y + z) + (z - x) (z + x) + (x - y) (x + y) \\ = y^{2} - z^{2} + z^{2} - x^{2} + x^{2} - y^{2} = 0 \end{matrix}$
Also it is orthogonal to $x \hat{i} + y \hat{j} + z \hat{k}$

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