Question

$\mathbf{a},\mathbf{b}$ and c are three vectors with magnitude $|\mathbf{a}| = 4,$ $|\mathbf{b}| = 4,$ $|\mathbf{c}| = 2$ and such that $\mathbf{a}$ is perpendicular to $(\mathbf{b} + \mathbf{c}),\mathbf{b}$ is perpendicular to $(\mathbf{c} + \mathbf{a})$ and $\mathbf{c}$ is perpendicular to $(\mathbf{a} + \mathbf{b}).$ It follows that $|\mathbf{a} + \mathbf{b} + \mathbf{c}|$ is equal to

• A. 9
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (B)

6

Answer 2

Here |a|=4; $|\mathbf{b}| = 4;|\mathbf{c}| = 2$

and $\mathbf{a}.(\mathbf{b} + \mathbf{c}) = 0 \Rightarrow \mathbf{a}.\mathbf{b} + \mathbf{a}.\mathbf{c} = 0$ .....(i)

$\mathbf{b}.(\mathbf{c} + \mathbf{a}) = 0 \Rightarrow \mathbf{b}.\mathbf{c} + \mathbf{b}.\mathbf{a} = 0$ .....(ii)

$\mathbf{c}.(\mathbf{a} + \mathbf{b}) = 0 \Rightarrow \mathbf{c}.\mathbf{a} + \mathbf{c}.\mathbf{b} = 0$ .....(iii)

Adding (i), (ii) and (iii), we get, $2\lbrack\mathbf{a}.\mathbf{b} + \mathbf{b}.\mathbf{c} + \mathbf{c}.\mathbf{a}\rbrack = 0$

∴ $|\mathbf{a} + \mathbf{b} + \mathbf{c}| = \sqrt{|\mathbf{a}|^{2} + |\mathbf{b}|^{2} + |\mathbf{c}|^{2} + 2(\mathbf{a}.\mathbf{b} + \mathbf{b}.\mathbf{c} + \mathbf{c}.\mathbf{a})}$

$= \sqrt{|\mathbf{a}|^{2} + |\mathbf{b}|^{2} + |\mathbf{c}|^{2}}$=$\sqrt{16 + 16 + 4}$

⇒ $|\mathbf{a} + \mathbf{b} + \mathbf{c}| = 6$.