Let a , b and c be three unit vectors such that |a-b|^ 2 +|a-c|^ … - Vidyadip

MCQ

Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$ Then $|\vec{a}+2 \vec{b}|^{2}+|\vec{a}+2 \vec{c}|^{2}$ is equal to

A
$1$
✓
$2$
C
$4$
D
$6$

✓

Answer

Correct option: B.

$2$

b
$|\vec{a}|=|\vec{b}|=|\vec{c}|=1$

$|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}=8$

$\Rightarrow \quad|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}+|\vec{a}|^{2}+|\vec{c}|^{2}-2 \vec{a} \cdot \vec{c}=8$

$\Rightarrow \quad 4-2(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})=8$

$\Rightarrow \quad \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=-2$

$|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{c}}|^{2}$

$=\left|\mathrm{a}^{2}\right|+4|\overrightarrow{\mathrm{b}}|^{2}+4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{a}}|^{2}+4|\overrightarrow{\mathrm{c}}|^{2}+4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$

$=10+4(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})$

$=10-8$

$=\sqrt{2}$

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