First slide

Vector triple product

Question

Let $\vec{a} and \vec{b}$ be two non-zero perpendicular vectors. A vector $\vec{r}$ satisfying the equation $\vec{r} \times \vec{b} = \vec{a}$ can be

Moderate

A

$\vec{b} - \frac{\vec{a} \times \vec{b}}{| \vec{b} |^{2}}$
B

$2 \vec{b} - \frac{\vec{a} \times \vec{b}}{| \vec{b} |^{2}}$
C

$| \vec{a} | \vec{b} - \frac{\vec{a} \times \vec{b}}{| \vec{b} |^{2}}$
D

$| \vec{b} | \vec{b} - \frac{\vec{a} \times \vec{b}}{| \vec{b} |^{2}}$

Solution

$Since \vec{a}, \vec{b} and \vec{a} \times \vec{b} are non-coplanar,$
$\begin{array}{ll} \vec{r} = x \vec{a} + y \vec{b} + z (\vec{a} \times \vec{b}) \\ ∴ \vec{r} \times \vec{b} = \vec{a} \\ \Rightarrow x \vec{a} \times \vec{b} + z {(\vec{a} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{a}} = \vec{a} \\ or - (1 + z | \vec{b} |^{2}) \vec{a} + x \vec{a} \times \vec{b} = 0 \end{array}$
$(since \vec{a} \cdot \vec{b} = 0)$
$∴ x = 0 and z = - \frac{1}{| \vec{b} |^{2}}$
$Thus, \vec{r} = y \vec{b} - \frac{\vec{a} \times \vec{b}}{| \vec{b} |^{2}},$ where y is the parameter.

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