If $\to a, \to b$ and $\to c$ are unit coplanar vectors then the scalar triple product $[2 \to a - \to b 2 \to b - \to c 2 \to c - \to a]$ is equal to

0

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1

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- \sqrt{3}

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\sqrt{3}

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Solution

The correct option is A $0$
Given that $[2 \to a - \to b 2 \to b - \to c 2 \to c - \to a]$

$= ∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & 0 0 & 2 & - 1 - 1 & 0 & 2 \end{matrix} ∣ ∣ ∣ ∣ [\to a \to b \to c]$
$= [2 (4 - 0) + 1 (0 - 1) + 0] [\to a \to b \to c]$
$= 7 [\to a \to b \to c]$
$= 0$ since $[\to a \to b \to c] = 0$ as $\to a, \to b, \to c$ are coplanar.

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Similar questions

If the scalar triple product of 3 vectors $\to a, \to b a n d \to c$ is 0 , then they are coplanar.

\to a, \to b, \to c

\frac{π}{3}

[\to a \to b \to c]

\to a, \to b, \to c

{\to b}_{1} = \to b - \frac{\to b \cdot \to a}{| \to a |^{2}} \to a,

{\to b}_{2} = \to b + \frac{\to b \cdot \to a}{| \to a |^{2}} \to a

{\to c}_{1} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1},

{\to c}_{2} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a - \frac{{\to b}_{1} \cdot \to c}{| {\to b}_{1} |^{2}} {\to b}_{1},

{\to c}_{3} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1}, {\to c}_{4} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a - \frac{\to b \cdot \to c}{| \to b |^{2}} {\to b}_{1}

\to a, \to b

\to c

| \to a - \to b |^{2} + | \to b - \to c |^{2} + | \to c - \to a |^{2} = 9

| 2 \to a + 5 \to b + 5 \to c |

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

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