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Cross Product of Two Vectors

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Question

Let the vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ such that $|\vec{a}| = 2$ , $|\vec{b}| = 4$ and $|\vec{c}| = 4$ . If the projection of vector $b$ on vector $a$ is equal to the projection of vector $c$ on vector $a$ and $b$ is perpendicular to vector $c$ , then the value of $|\vec{a} + \vec{b} - \vec{c}|$ is :

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Solution

Finding the value of $|\vec{a} + \vec{b} - \vec{c}|$ :
Given the vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ such that $|\vec{a}| = 2$ , $|\vec{b}| = 4$ and $|\vec{c}| = 4$ .
According to the given condition that the projection of vector $b$ on vector $c$ is equal to the projection of vector $c$ on vector $a$ ,
$\Rightarrow \vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{a}$
Since vector $b$ is perpendicular to vector $c$ .
$\Rightarrow \vec{b} \cdot \vec{c} = 0$
We know that, ${|\vec{a} + \vec{b} - \vec{c}|}^{2} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} - 2 (\vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c})$ from the property of vector.
Now substitute the value and compute the value of ${|\vec{a} + \vec{b} - \vec{c}|}^{2}$ .
$\begin{array}{rcl} {|\vec{a} + \vec{b} - \vec{c}|}^{2} & = & {(2)}^{2} + {(4)}^{2} + {(4)}^{2} - 2 (\vec{a} \cdot \vec{b} - 0 - \vec{a} \cdot \vec{b}) \\ = & 4 + 16 + 16 - 2 (\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b}) \\ = & 36 - 2 (0) \\ = & 36 \end{array}$
Take square root on both side,
$\begin{array}{rcl} \sqrt{{(|\vec{a} + \vec{b} - \vec{c}|)}^{2}} & = & \sqrt{36} \\ |\vec{a} + \vec{b} - \vec{c}| & = & 6 \end{array}$
Therefore, the value of $|\vec{a} + \vec{b} - \vec{c}|$ is $6$ .

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