Question

If the non-zero vectors a and b are perpendicular to each other, then the solution of the equation $\mathbf{r} \times \mathbf{a} = \mathbf{b}$ is given by

• A. $\mathbf{r} = x\mathbf{a} + \frac{1}{\mathbf{a}.\mathbf{a}}(\mathbf{a} \times \mathbf{b})$
• B. $\mathbf{r} = x\mathbf{b} - \frac{1}{\mathbf{b}.\mathbf{b}}(\mathbf{a} \times \mathbf{b})$
• C. $\mathbf{r} = x\mathbf{a} \times \mathbf{b}$
• D. $\mathbf{r} = x\mathbf{b} \times \mathbf{a}$

Answer 1

Answer: (A)

$\mathbf{r} = x\mathbf{a} + \frac{1}{\mathbf{a}.\mathbf{a}}(\mathbf{a} \times \mathbf{b})$

Answer 2

Since $\mathbf{a},\mathbf{b}$ and $\mathbf{a} \times \mathbf{b}$ are non-coplanar,

hence $\mathbf{r} = x\mathbf{a} + y\mathbf{b} + z(\mathbf{a} \times \mathbf{b})$ for some scalars $x,y$ and $z.$

Now, $\mathbf{b} = \mathbf{r} \times \mathbf{a} = \left\{ x\mathbf{a} + y\mathbf{b} + z(\mathbf{a} \times \mathbf{b}) \right\} \times \mathbf{a}$

$= y(\mathbf{b} \times \mathbf{a}) + z\lbrack(\mathbf{a} \times \mathbf{b}) \times \mathbf{a}\rbrack = - y(\mathbf{a} \times \mathbf{b}) - z\lbrack\mathbf{a} \times (\mathbf{a} \times \mathbf{b})\rbrack$

$= - y(\mathbf{a} \times \mathbf{b}) - z\lbrack(\mathbf{a}.\mathbf{b})\mathbf{a} - (\mathbf{a}.\mathbf{a})\mathbf{b}\rbrack$

$= - y(\mathbf{a} \times \mathbf{b}) + z(\mathbf{a}.\mathbf{a})\mathbf{b}$, $\left\{ \because\mathbf{a}.\mathbf{b} = 0 \right\}$

$\Rightarrow y = 0$ and $z = \frac{1}{(\mathbf{a}.\mathbf{a})} \Rightarrow \mathbf{r} = x\mathbf{a} + \frac{1}{\mathbf{a}.\mathbf{a}}(\mathbf{a} \times \mathbf{b})$.