# Question #420ec

Aug 14, 2017

Given

• $\vec{p} = \hat{i} + 2 \hat{j} - 2 \hat{k}$

• $\vec{q} = 2 \hat{i} - \hat{j} + 2 \hat{k}$

we are to find $\vec{m} \mathmr{and} \vec{n}$ satisfying conditions

• (a) $\vec{m}$ is perpendicular to $\vec{p}$

• (b) $\vec{n}$ is parallel to $\vec{p}$

• (c) $\vec{m} + \vec{n} = \vec{q}$

By condition (b)$\vec{n}$ is parallel to $\vec{p}$

So

• We can write
$\vec{n} = \alpha \cdot \vec{p}$,where $\alpha$ is a constant.

$\vec{n} = \alpha \vec{p} = \alpha \hat{i} + 2 \alpha \hat{j} - 2 \alpha \hat{k}$

By condition (c)

$\vec{m} + \vec{n} = \vec{q}$

$\implies \vec{m} = \vec{q} - \vec{n}$

$\implies \vec{m} = \left(2 - \alpha\right) \hat{i} - \left(1 + 2 \alpha\right) \hat{j} + \left(2 + 2 \alpha\right) \hat{k}$

Now by condition (a) $\vec{m}$ is perpendicular to $\vec{p}$

So $\vec{m} \cdot \vec{p} = 0$

$\implies 1 \times \left(2 - \alpha\right) - 2 \left(1 + 2 \alpha\right) - 2 \left(2 + 2 \alpha\right) = 0$

$\implies 2 - \alpha - 2 - 4 \alpha - 4 - 4 \alpha = 0$

$\implies \alpha = - \frac{4}{9}$

Now
$\vec{n} = \alpha \hat{i} + 2 \alpha \hat{j} - 2 \alpha \hat{k}$

$\implies \vec{n} = - \frac{4}{9} \hat{i} - \frac{8}{9} \hat{j} + \frac{8}{9} \hat{k}$

and

$\vec{m} = \left(2 - \alpha\right) \hat{i} - \left(1 + 2 \alpha\right) \hat{j} + \left(2 + 2 \alpha\right) \hat{k}$

$\implies \vec{m} = \left(2 + \frac{4}{9}\right) \hat{i} - \left(1 - \frac{8}{9}\right) \hat{j} + \left(2 - \frac{8}{9}\right) \hat{k}$

$\implies \vec{m} = \frac{22}{9} \hat{i} - \frac{1}{9} \hat{j} + \frac{10}{9} \hat{k}$