If a- b- c- are non-coplanar unit vectors such that a- b-c- - b-c-2- then the angle between a- and b- is

Since, a⃗× (b⃗×c⃗) = b⃗+c⃗/√(2) ⇒ (a⃗.c⃗)b⃗ (a⃗.b⃗)c⃗ = 1/√(2)b⃗ + 1/√(2)c⃗ On equating the coefficient of c⃗, we get a⃗.b⃗ = 1/√(2) ⇒ |a⃗||b⃗| cos θ = 1/ ...

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Byju's Answer

Standard XIII

Mathematics

Vector Triple Product

If a⃗, b⃗, c⃗...

Question

If $\to a, \to b, \to c$ are non-coplanar unit vectors such that $\to a \times (\to b \times \to c) = \frac{(\to b + \to c)}{\sqrt{2}},$ then the angle between $\to a$ and $\to b$ is

\frac{3 π}{4}

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\frac{π}{4}

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\frac{π}{2}

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π

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Solution

The correct option is A $\frac{3 π}{4}$
Since, $\to a \times (\to b \times \to c) = \frac{\to b + \to c}{\sqrt{2}}$
$\Rightarrow (\to a . \to c) \to b - (\to a . \to b) \to c = \frac{1}{\sqrt{2}} \to b + \frac{1}{\sqrt{2}} \to c$
On equating the coefficient of $\to c,$ we get
$\to a . \to b = - \frac{1}{\sqrt{2}} \Rightarrow | \to a | | \to b | c o s θ = - \frac{1}{\sqrt{2}} ∴ c o s θ = - \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{3 π}{4}$

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If →a,→b,→c are non-coplanar unit vectors such that →a×(→b×→c)=(→b+→c)√2, then the angle between →a and →b is

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