Q117 / 170 · Manipal (MET) 2021 · Mathematics
$[\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]$ is equal to
- A$[\vec{a} \vec{b} \vec{c}]$
- B$\Sigma(\vec{a} \cdot \vec{b}) \vec{c}$
- Correct answerC$2[\vec{a} \vec{b} \vec{c}]$
- D$|\vec{a}||\vec{b}||\vec{c}|$
Solution
$\begin{aligned} & {[\vec{a}+\vec{b} \vec{b}=+\vec{a}} \\ & =(\vec{a}+\vec{b}) \cdot[(\vec{b}+\vec{c}) \times(\vec{c}+\vec{a})] \\ & =(\vec{a}+\vec{b}) \cdot[(\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+0+\vec{c} \times \vec{a})]\end{aligned}$
$\begin{aligned} & =[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]+0+0+0+0+[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{a}}] \\ & =[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]+[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \\ & =2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]\end{aligned}$
Hence, the correct answer is option: 'C'.