Q108 / 180 · COMEDK 2015 · Mathematics

If a, be are three non-zero vectors such that each one of them are perpendicular to the sum of the other two vectors, then the value of $\mid \mathbf{a}+\mathbf{b}+\mathbf{c}^{2}$ is

  • Correct answer
    A
    $|a|^{2}+|b|^{2}+\mid q^{2}$
  • B
    $\mid$ à $+|b|+|\mathbb{d}|$
  • C
    $2\left(|\mathbf{a}|^{2}+|b|^{2}+\mid d^{2}\right)$
  • D
    $\frac{1}{2}\left(|a|^{2}+|b|^{2}+\mid d^{2}\right)$

Solution

According to the question, $a \cdot(b+c)=0, b \cdot(\mathfrak{c}+\mathfrak{a})=0, \mathfrak{c} \cdot(\mathbf{a}+\mathbf{b})=0$ $\Rightarrow \quad \quad$ a $\cdot b+a \cdot c=0 \quad \text{...(i)}$ b. $c+b \cdot a=0 \quad \text{...(ii)}$ $c \cdot \mathbf{a}+c \cdot b=0 \quad \text{...(iii)}$ Adding Eqs. (i), (ii) and (iii), we get $2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} . \mathbf{a})=0 \quad \text{...(iv)}$ Now, $\mid a+b+c^{2}$ $=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+\mid \mathbf{c}^{2}+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})$ $=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+|\mathbb{c}|^{2} \quad$ [using Eq. (iv)]

Hence, the correct answer is option: 'A'.

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