MCQ
If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors having magnitudes $1,2,3$ respectively, then $[\vec{a}+\vec{b}+\vec{c} \vec{b}-\vec{a} \vec{c}]=$
- A0
- B6
- ✓12
- D18
✓
Answer
Correct option: C.
12
(c) : Given, $|\vec{a}|=1,|\vec{b}|=2,|\vec{c}|=3$
$\because \quad \vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors
$
\because \quad \vec{a} \cdot \vec{b}=0=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}
$
Now, $[\vec{a}+\vec{b}+\vec{c} \quad \vec{b}-\vec{a} \vec{c}]=[(\vec{a}+\vec{b}+\vec{c}) \times(\vec{b}-\vec{a})] \cdot \vec{c}$
$
=[\vec{a} \times \vec{b}-\vec{a} \times \vec{a}+\vec{b} \times \vec{b}-\vec{b} \times \vec{a}+\vec{c} \times \vec{b}-\vec{c} \times \vec{a}] \cdot \vec{c}
$
$
\begin{aligned}
& =[\vec{a} \times \vec{b}-0+0+\vec{a} \times \vec{b}+\vec{c} \times \vec{b}-\vec{c} \times \vec{a}] \cdot \vec{c} \\
& =[2(\vec{a} \times \vec{b})+\vec{c} \times \vec{b}-\vec{c} \times \vec{a}] \cdot \vec{c}=2(\vec{a} \times \vec{b}) \cdot \vec{c}+0-0 \\
& =2[\vec{a} \vec{b} \quad \vec{c}]=2 \vec{a} \cdot(\vec{b} \times \vec{c})=2|\vec{a}||\vec{b} \times \vec{c}| \cos 0^{\circ} \\
& =2(1)(2)(3)=12
\end{aligned}
$
$\because \quad \vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors
$
\because \quad \vec{a} \cdot \vec{b}=0=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}
$
Now, $[\vec{a}+\vec{b}+\vec{c} \quad \vec{b}-\vec{a} \vec{c}]=[(\vec{a}+\vec{b}+\vec{c}) \times(\vec{b}-\vec{a})] \cdot \vec{c}$
$
=[\vec{a} \times \vec{b}-\vec{a} \times \vec{a}+\vec{b} \times \vec{b}-\vec{b} \times \vec{a}+\vec{c} \times \vec{b}-\vec{c} \times \vec{a}] \cdot \vec{c}
$
$
\begin{aligned}
& =[\vec{a} \times \vec{b}-0+0+\vec{a} \times \vec{b}+\vec{c} \times \vec{b}-\vec{c} \times \vec{a}] \cdot \vec{c} \\
& =[2(\vec{a} \times \vec{b})+\vec{c} \times \vec{b}-\vec{c} \times \vec{a}] \cdot \vec{c}=2(\vec{a} \times \vec{b}) \cdot \vec{c}+0-0 \\
& =2[\vec{a} \vec{b} \quad \vec{c}]=2 \vec{a} \cdot(\vec{b} \times \vec{c})=2|\vec{a}||\vec{b} \times \vec{c}| \cos 0^{\circ} \\
& =2(1)(2)(3)=12
\end{aligned}
$
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