If a- and b- are unequal vectors such that a-b- -b-a- -2 a-b-a-b- then the angle - between a- and b- is

The correct options are B π2 D π(a b)×[(b+a)×(2a+b)]=a+b or {(a b)·(2a+b)}(b+a) {(a b)·(b+a)}(2a+b)=b+a or (2 a·b 1)(b+a)=b+a ⇒ either b+a=0 or 1 a·b=1 ⇒ eith ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

If a⃗ and b⃗ ...

Question

If $\to a$ and $\to b$ are unequal vectors such that $(\to a - \to b) \times [(\to b + \to a) \times (2 \to a + \to b)] = \to a + \to b$ , then the angle $θ$ between $\to a$ and $\to b$ is

0

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\frac{π}{2}

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\frac{π}{4}

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π

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Solution

The correct options are
B $\frac{π}{2}$
D $π$
$(\to a - \to b) \times [(\to b + \to a) \times (2 \to a + \to b)] = \to a + \to b$

or ${(\to a - \to b) \cdot (2 \to a + \to b)} (\to b + \to a)$ $- {(\to a - \to b) \cdot (\to b + \to a)} (2 \to a + \to b) = \to b + \to a$

or $(2 - \to a \cdot \to b - 1) (\to b + \to a) = \to b + \to a$
$\Rightarrow$ either $\to b + \to a = \to 0$ or $1 - \to a \cdot \to b = 1$
$\Rightarrow$ either $\to b = - \to a$ or $\to a \cdot \to b = 0$
$\Rightarrow$ either $θ = π$ or $θ = \frac{π}{2}$

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If →a and →b are unequal vectors such that (→a−→b)×[(→b+→a)×(2→a+→b)]=→a+→b, then the angle θ between →a and →b is

If $\to a$ and $\to b$ are unequal vectors such that $(\to a - \to b) \times [(\to b + \to a) \times (2 \to a + \to b)] = \to a + \to b$ , then the angle $θ$ between $\to a$ and $\to b$ is