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If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to
the plane of $\vec{b}, \vec{c}$; and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$
Then, what is $|\vec{a}+\vec{b}+\vec{c}| ?$
Options:
the plane of $\vec{b}, \vec{c}$; and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$
Then, what is $|\vec{a}+\vec{b}+\vec{c}| ?$
Solution:
1818 Upvotes
Verified Answer
The correct answer is:
2
As given: a is perpendicular to $\overrightarrow{\mathrm{b}}$ and $\mathrm{c}$ $\Rightarrow \vec{a} \cdot \vec{b}=0 \& \overrightarrow{a \cdot c}=0$
and angle between $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}=\frac{\pi}{3}$
$\therefore \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \frac{\pi}{3}=1.1 \cdot \frac{1}{2}$
$=\frac{1}{2} \quad(\because \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are unit vectors)
Now, $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+|\vec{c}|^{2}$
$+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a})$
$=1+1+1+2 \cdot\left(0+\frac{1}{2}+0\right)$
$=1+1+1+1=4$
$\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|=2$
and angle between $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}=\frac{\pi}{3}$
$\therefore \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \frac{\pi}{3}=1.1 \cdot \frac{1}{2}$
$=\frac{1}{2} \quad(\because \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are unit vectors)
Now, $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+|\vec{c}|^{2}$
$+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a})$
$=1+1+1+2 \cdot\left(0+\frac{1}{2}+0\right)$
$=1+1+1+1=4$
$\Rightarrow|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|=2$
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