Question

Let $\vec{a}, \vec{b}, \vec{c}$
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
• C. 16
• D. 18

Answer 1

Answer: (C)

16

Answer 2

$|\vec{a}| |\vec{b}| |\vec{c}| = 14$
$\vec{a} \land \vec{b} = \vec{b} \land \vec{c} = \vec{c} \land \vec{a} = \theta = \frac{2\pi}{3}$
$\vec{a} \cdot \vec{b} = -\frac{1}{2} |\vec{a}| |\vec{b}|$
$\vec{b} \cdot \vec{c} = -\frac{1}{2} |\vec{b}| |\vec{c}|$
$\vec{c} \cdot \vec{a} = -\frac{1}{2} |\vec{c}| |\vec{a}|$
Now,
$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168\ \ \ ....(i)$
$(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})|\vec{b}|^2$
$= \frac{1}{4} |\vec{b}|^2 |\vec{a}| |\vec{c}| + \frac{1}{2} |\vec{a}| |\vec{b}|^2 |\vec{c}|$
$= \frac{3}{4} |\vec{a}| |\vec{b}|^2 |\vec{c}|$ $... (ii)$
Similarly $(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4} |\vec{a}| |\vec{b}| |\vec{c}|^2$ $... (iii)$
$(\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4} |\vec{a}|^2 |\vec{b}| |\vec{c}|$ $... (iv)$
Substitute (ii),(iii),(iv) in (i)
$\frac{3}{4} |\vec{a}| |\vec{b}| |\vec{c}| \left[ |\vec{a}| + |\vec{b}| + |\vec{c}| \right] = 168$
$\frac{3}{4} \times 14 \left[ |\vec{a}| + |\vec{b}| + |\vec{c}| \right] = 168$
$|\vec{a}| + |\vec{b}| + |\vec{c}| = 16$
So, the correct answer is (C) : 16