If a- - b- - c- are unit vectors such that a- - b- - 0 - a-c- and the angle between b- and c- is - - 3 - then the value to - a-b - a-c- is -

The correct option is A 1 We have #160; | a #160; ×b #160; a #160; ×c| = | a| ×| b #160; c| = | a|| b #160; c|sinθ = ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

If a⃗ , b⃗ ...

Question

If $\to a, \to b, \to c$ are unit vectors such that $\to a . \to b = 0 = \to a . \to c$ and the angle between $\to b$ and $\to c$ is $π / 3$ , then the value to $∣ ∣ ∣ \to a \times \to b - \to a \times \to c ∣ ∣ ∣$ is ?

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Solution

The correct option is A $1$
We have
$∣ ∣ ∣ \to a \times \to b - \to a \times \to c ∣ ∣ ∣$
$= ∣ ∣ \to a ∣ ∣ \times ∣ ∣ ∣ \to b - \to c ∣ ∣ ∣$
$= ∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to b - \to c ∣ ∣ ∣ sin θ$
$= ∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to b - \to c ∣ ∣ ∣$
$= 1$

We have
$\to a (\to b - \to c) = 0$
$\to a ⊥ \to b - \to c \Rightarrow θ = 90^{\circ}$
Now
${∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} = {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} - 2 \to b - \to c$
${∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} = 1 + 1 - 2 \times 1 \times 1 \times \frac{1}{2}$
${∣ ∣ ∣ \to b - \to c ∣ ∣ ∣}^{2} = 1$
$∴ ∣ ∣ ∣ \to b - \to c ∣ ∣ ∣ = 1$
Hence, the option $B$ is the correct answer.

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If →a,→b,→c are unit vectors such that →a.→b=0=→a.→c and the angle between →b and →c is π/3 , then the value to∣∣∣→a×→b−→a×→c∣∣∣ is ?

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