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If two unit vectors $\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{q}}$ make an angle $\frac{\pi}{3}$ with each
other, what is the magnitude of $\vec{p}-\frac{1}{2} \vec{q}$ ?
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other, what is the magnitude of $\vec{p}-\frac{1}{2} \vec{q}$ ?
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2}$
If $\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{q}}$ are unit vectors which make an angle $\frac{\pi}{3}$ with each other.
Then, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}=|\overrightarrow{\mathrm{p}}||\overrightarrow{\mathrm{q}}| \cos \frac{\pi}{3}=\frac{1}{2}$
Now, $\left|\vec{p}-\frac{1}{2} \vec{q}\right|^{2}=|\vec{p}|^{2}+\frac{1}{4}|\vec{q}|^{2}-\frac{2}{2} \vec{p} \cdot \vec{q}$
$=1+\frac{1}{4}-\frac{1}{2} \quad[$ since $|\overrightarrow{\mathrm{p}}|=|\overrightarrow{\mathrm{q}}|=1]$
$=\frac{5}{4}-\frac{1}{2}=\frac{5-2}{4}=\frac{3}{4}$
So, $\left|\vec{p}-\frac{1}{2} \vec{q}\right|=\frac{\sqrt{3}}{2}$
Then, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}=|\overrightarrow{\mathrm{p}}||\overrightarrow{\mathrm{q}}| \cos \frac{\pi}{3}=\frac{1}{2}$
Now, $\left|\vec{p}-\frac{1}{2} \vec{q}\right|^{2}=|\vec{p}|^{2}+\frac{1}{4}|\vec{q}|^{2}-\frac{2}{2} \vec{p} \cdot \vec{q}$
$=1+\frac{1}{4}-\frac{1}{2} \quad[$ since $|\overrightarrow{\mathrm{p}}|=|\overrightarrow{\mathrm{q}}|=1]$
$=\frac{5}{4}-\frac{1}{2}=\frac{5-2}{4}=\frac{3}{4}$
So, $\left|\vec{p}-\frac{1}{2} \vec{q}\right|=\frac{\sqrt{3}}{2}$
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