Let a , b , c be three coplanar concurrent vectors such that angl… - Vidyadip

MCQ

Let $\overrightarrow{ a }, \overrightarrow{ b }, \overrightarrow{ c }$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is $14$ and $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168$ then $|\vec{a}|+|\vec{b}|+|\vec{c}|$ is equal to.

A
$10$
B
$14$
✓
$16$
D
$18$

✓

Answer

Correct option: C.

$16$

c
$|\vec{a}\|\vec{b}\| \vec{c}|=14$

$\overrightarrow{ a }^{\wedge} \overrightarrow{ b }=\overrightarrow{ b } \wedge \overrightarrow{ c }=\overrightarrow{ c }^{\wedge} \overrightarrow{ a }=\theta=\frac{2 \pi}{3}$

So, $\vec{a} \cdot \vec{b}=-\frac{1}{2} a b, \vec{b} \cdot \vec{c}=-\frac{1}{2} b c, \vec{a} \cdot \vec{c} .=-\frac{1}{2} a c$ (let)

$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})-(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b})$

$=\frac{1}{4} a b^{2} c+\frac{1}{2} a b^{2} c=\frac{3}{4} a b^{2} c$

Similarly

$(\overrightarrow{ b } \times \overrightarrow{ c }) \cdot(\overrightarrow{ c } \times \overrightarrow{ a })=\frac{3}{4} abc ^{2}$

$(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=\frac{3}{4} a^{2} b c$

$168=\frac{3}{4} a b c(a+b+c)$

So, $(a+b+c)=16$

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