If two vectors $\bar{a}$ and $\bar{b}$ which are perpendicular to each other are such that $|\bar{a}| = 8$ and $| \bar{b} | = 3, then | \bar{a} - 2 \bar{b} | =$

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Detailed Solution

$|\bar{a}| = 8$ and $| \bar{b} | = 3$

$Since \vec{a} and \vec{b} are perpendicular$

so, $\vec{a} \cdot \vec{b} = 0$

$\begin{aligned} | \vec{a} - 2 \vec{b} |^{2} = (\vec{a} - 2 \vec{b}) \cdot (\vec{a} - 2 \vec{b}) \\ = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} - 2 \vec{b} \cdot \vec{a} + 4 \vec{b} \cdot \vec{b} \\ = | \vec{a} |^{2} - 2 \vec{a} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} + 4 | \vec{b} |^{2} (\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}, \vec{a} \cdot \vec{a} = | \vec{a} |^{2}) \\ = (8)^{2} - 4 \vec{a} \cdot \vec{b} + 4 (3)^{2} \\ = 64 - 4 \times 0 + 4 \times 9 \\ = 64 + 36 \\ = 100 \\ | \vec{a} - 2 \vec{b} | = \sqrt{100} \\ | \vec{a} - 2 \vec{b} | = 10 \end{aligned}$