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Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be unit vectors such that $\mathbf{a} \cdot \mathbf{b}=0=\mathbf{a} \cdot \mathbf{c}$ and the acute angle between $\mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$, then $|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|$ is equal to
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The correct answer is:
$1$
Given, $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$
and $\mathbf{a} \cdot \mathbf{b}=0=\mathbf{a} \cdot \mathbf{c}$ angle between $\mathbf{b}$ and $\mathbf{c}=\frac{\pi}{3}$
To find, $|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|$
Given that, $\mathbf{a} \cdot \mathbf{b}=0$ and $\mathbf{a} \cdot \mathbf{c}=0$
$\Rightarrow \quad \mathbf{a} \cdot(\mathbf{b}-\mathbf{c})=0$
$\Rightarrow$ The angle between $\mathbf{a}$ and $(\mathbf{b}-\mathbf{c})$ is $\frac{\pi}{2}$.
Now, $|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|=|\mathbf{a} \times(\mathbf{b}-\mathbf{c})|$
$=|\mathbf{a}||\mathbf{b}-\mathbf{c}| \sin 90^{\circ}$ ...(i)
$|\mathbf{b}-\mathbf{c}|^2=|\mathbf{b}|^2+|\mathbf{c}|^2-2 \mathbf{b} \cdot \mathbf{c}$
$\begin{aligned} & =1+1-2\left(|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3}\right) \\ & =2-2\left(1 \cdot 1 \cdot \frac{1}{2}\right)\end{aligned}$
$|\mathbf{b}-\mathbf{c}|^2=1$
$\Rightarrow|\mathbf{b}-\mathbf{c}|=1$
From Eq. (i), we get
$|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|=1$
and $\mathbf{a} \cdot \mathbf{b}=0=\mathbf{a} \cdot \mathbf{c}$ angle between $\mathbf{b}$ and $\mathbf{c}=\frac{\pi}{3}$
To find, $|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|$
Given that, $\mathbf{a} \cdot \mathbf{b}=0$ and $\mathbf{a} \cdot \mathbf{c}=0$
$\Rightarrow \quad \mathbf{a} \cdot(\mathbf{b}-\mathbf{c})=0$
$\Rightarrow$ The angle between $\mathbf{a}$ and $(\mathbf{b}-\mathbf{c})$ is $\frac{\pi}{2}$.
Now, $|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|=|\mathbf{a} \times(\mathbf{b}-\mathbf{c})|$
$=|\mathbf{a}||\mathbf{b}-\mathbf{c}| \sin 90^{\circ}$ ...(i)
$|\mathbf{b}-\mathbf{c}|^2=|\mathbf{b}|^2+|\mathbf{c}|^2-2 \mathbf{b} \cdot \mathbf{c}$
$\begin{aligned} & =1+1-2\left(|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3}\right) \\ & =2-2\left(1 \cdot 1 \cdot \frac{1}{2}\right)\end{aligned}$
$|\mathbf{b}-\mathbf{c}|^2=1$
$\Rightarrow|\mathbf{b}-\mathbf{c}|=1$
From Eq. (i), we get
$|\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}|=1$
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