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Distance Formula

If a,b,c ar...

Question

If $\to a, \to b, \to c$ are unit vectors equally inclined to each other at an angle $\frac{π}{3}$ , then $[\to b + \to c, \to c + \to a, \to a + \to b] =$

A

1

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B

\frac{1}{\sqrt{2}}

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C

\frac{1}{2}

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D

\sqrt{2}

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Solution

The correct option is C $\sqrt{2}$
${[\to a \to b \to c]}^{2} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a . \to a & \to a . \to b & \to a . \to c \to b . \to a & \to b . \to b & \to b . \to c \to c . \to a & \to c . \to b & \to c . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2} \frac{1}{2} & 1 & \frac{1}{2} \frac{1}{2} & \frac{1}{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$= 1 (1 - \frac{1}{4}) - \frac{1}{2} (\frac{1}{2} - \frac{1}{4}) + \frac{1}{2} (\frac{1}{4} - \frac{1}{2})$
$= \frac{3}{4} - \frac{1}{2} (\frac{1}{4}) + \frac{1}{2} (- \frac{1}{4})$
$= \frac{6 - 1 - 1}{8} = \frac{4}{8} = \frac{1}{2}$
$[\to a \to b \to c] = \frac{1}{\sqrt{2}}$
$\Rightarrow [\to b + \to c, \to c + \to a, \to a + \to b]$
$= ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦ [\to a, \to b, \to c]$
$= [0 - 1 (0 - 1) + 1 (1 - 0)] [\to a, \to b, \to c]$
$= 2 [\to a \to b \to c]$
$= 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

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