Given that P- 12- Q- 5 and R- 13 also P-Q- R- then the angle between P- and Q- will be-

The correct option is B π/2 It can be seen that P^2+Q^2=R^2 12^2+5^2=13^2 But P⃗+Q⃗=R⃗ , which implies P^2+Q^2+2PQ.cosθ =R^2 Thus, ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

Given that ...

Question

Given that $P = 12, Q = 5$ and $R = 13$ also $\to P + \to Q = \to R,$ then the angle between $\to P$ and $\to Q$ will be:

π

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\frac{π}{2}

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zero

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\frac{π}{4}

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Solution

The correct option is B $\frac{π}{2}$
It can be seen that $P^{2} + Q^{2} = R^{2}$
$12^{2} + 5^{2} = 13^{2}$
But $\to P + \to Q = \to R$ , which implies $P^{2} + Q^{2} + 2 P Q . c o s θ = R^{2}$
Thus, $c o s θ = 0$ or $θ = \frac{π}{2}$

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Given that P=12,Q=5 and R=13 also →P+→Q=→R, then the angle between →P and →Q will be:

The correct option is B π2It can be seen that P2+Q2=R2122+52=132But →P+→Q=→R, which implies P2+Q2+2PQ.cosθ=R2Thus, cosθ=0 or θ=π2

Given that $P = 12, Q = 5$ and $R = 13$ also $\to P + \to Q = \to R,$ then the angle between $\to P$ and $\to Q$ will be:

The correct option is B $\frac{π}{2}$
It can be seen that $P^{2} + Q^{2} = R^{2}$
$12^{2} + 5^{2} = 13^{2}$
But $\to P + \to Q = \to R$ , which implies $P^{2} + Q^{2} + 2 P Q . c o s θ = R^{2}$
Thus, $c o s θ = 0$ or $θ = \frac{π}{2}$