If P→×Q→ $$=Q$$→×P→, the angle between P→ and Q→ is θ(00$$<θ<$$3600). The value of ‘θ’ will be ——

P→×Q→$$=Q$$→×P→⇒P→×Q→$$=$$−(P$$→×Q→$$)⇒$$2(P$$→×Q→$$)=0$$⇒$$2PQsin$$⁡θ$$=0$$ This is true when either θ$$=00$$ or $$180$$∘ Given : $$0$$∘<θ<$$360$$∘∴θ$$=180$$∘

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If $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P},$ the angle between $\vec{P} a n d \vec{Q} i s θ (0^{0} < θ < 360^{0})$ . The value of ‘θ’ will be ——

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$\begin{array}{l} \vec{P} \times \vec{Q} = \vec{Q} \times \vec{P} \\ \Rightarrow \vec{P} \times \vec{Q} = - (\vec{P} \times \vec{Q}) \\ \Rightarrow 2 (\vec{P} \times \vec{Q}) = 0 \\ \Rightarrow 2 P Q \sin θ = 0 \\ This is true when either θ = 0^{0} or 180^{\circ} \\ Given : (0^{\circ} < θ < 360^{\circ}) \\ ∴ θ = 180^{\circ} \end{array}$

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Scalars and Vectors

Remember concepts with our Masterclasses.

If P→×Q→ =Q→×P→, the angle between P→ and Q→ is θ(00<θ<3600). The value of ‘θ’ will be ——

P→×Q→=Q→×P→⇒P→×Q→=−(P→×Q→)⇒2(P→×Q→)=0⇒2PQsin⁡θ=0 This is true when either θ=00 or 180∘ Given : 0∘<θ<360∘∴θ=180∘

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free study material

If $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P},$ the angle between $\vec{P} a n d \vec{Q} i s θ (0^{0} < θ < 360^{0})$ . The value of ‘θ’ will be ——