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Show that each of the given three vectors is a unit vector: $\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k}), \frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$. Also show that they are mutually perpendicular to each other.
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Verified Answer
Let $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})$, $\overrightarrow{\mathrm{c}}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$
Now, $|\vec{a}|=1,|\vec{b}|=1,|\vec{c}|=1$
$\therefore \quad \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are unit vectors.
Now, $\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0, \vec{c} \cdot \vec{a}=0$
$\Rightarrow \quad \vec{a} \perp \vec{b}, \vec{b} \perp \vec{c}$ and $\vec{c} \perp \vec{a}$
Hence, $\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular unit yectors.
Now, $|\vec{a}|=1,|\vec{b}|=1,|\vec{c}|=1$
$\therefore \quad \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are unit vectors.
Now, $\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0, \vec{c} \cdot \vec{a}=0$
$\Rightarrow \quad \vec{a} \perp \vec{b}, \vec{b} \perp \vec{c}$ and $\vec{c} \perp \vec{a}$
Hence, $\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular unit yectors.
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