The non-zero vectors a-b- and c- are related by a-8b- and c-7b- then the angle between a- and c- is

The correct option is A π Given, a⃗=8 b⃗ , c⃗= 7b⃗ Therefore, a⃗.c⃗=accosθ ⇒ 8b⃗.( 7b⃗)=8|b|(7|b|)cosθ ⇒ 56b⃗.b⃗=56b^2cosθ ...

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Standard XII

Physics

Vector Addition

The non-zero ...

Question

The non-zero vectors $\to a, \to b$ and $\to c$ are related by $\to a = 8 \to b$ and $\to c = - 7 \to b$ , then the angle between $\to a$ and $\to c$ is

\frac{π}{2}

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π

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0

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\frac{π}{4}

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Solution

The correct option is A $π$
Given, $\to a = 8 \to b$ , $\to c = - 7 \to b$
Therefore, $\to a . \to c = a c cos θ$
$\Rightarrow 8 \to b . (- 7 \to b) = 8 | b | (7 | b |) cos θ$
$\Rightarrow - 56 \to b . \to b = 56 b^{2} cos θ$
$\Rightarrow - b^{2} = b^{2} cos θ$
$\Rightarrow cos θ = - 1$
$\Rightarrow θ = π$

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The non-zero vectors →a,→b and →c are related by →a=8→b and →c=−7→b, then the angle between →a and →c is

The correct option is A πGiven, →a=8→b, →c=−7→bTherefore, →a.→c=accosθ⇒8→b.(−7→b)=8|b|(7|b|)cosθ⇒−56→b.→b=56b2cosθ⇒−b2=b2cosθ⇒cosθ=−1⇒θ=π

The non-zero vectors $\to a, \to b$ and $\to c$ are related by $\to a = 8 \to b$ and $\to c = - 7 \to b$ , then the angle between $\to a$ and $\to c$ is