If vector A→$$=$$$$cos$$ω$$ti^+$$$$sin$$ω$$tj^$$ and B→$$=$$$$cos$$ω$$t2i^+$$$$sin$$ω$$t2j^$$ are functions of time, then the value of t at which they are orthogonal to each other is

Question

If vector A→$$=$$$$cos$$ω$$ti^+$$$$sin$$ω$$tj^$$  and  B→$$=$$$$cos$$ω$$t2i^+$$$$sin$$ω$$t2j^$$  are functions of time,  then the value of t at which they are orthogonal to each other is

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Two vectors A→ and B→ are orthogonal to each other, if  their scalar product is zero i.e. A→·B→$$=0$$ .  Here, A→$$=$$$$cos$$ω$$ti^+$$$$sin$$ω$$tj^$$  and   B→$$=$$$$cos$$ω$$t2i^+$$$$sin$$ω$$t2j^$$∴A→·B→$$=($$$$cos$$ω$$ti^+$$$$sin$$ω$$tj^)$$·$$cos$$ω$$t2i^+$$$$sin$$ω$$t2j^=$$$$cos$$ωtcosω$$t2+$$$$sin$$ωtsinω$$t2($$∵$$i^$$·$$i^=j^$$·$$j^=1$$ and $$i^$$·$$j^=j^$$·$$i^=0)=$$$$cos$$ω$$t-$$ω$$t2($$∵$$cos$$(A-B)=cosAcosB+sinAsinB) But A→·B→$$=0$$ (as$$ A→ and B→ are orthogonal to each $$other) ∴  $$cos$$ω$$t-$$ω$$t2=0cos$$ω$$t-$$ω$$t2=$$$$cos$$$$π$$$$2$$ or ω$$t-$$ω$$t2=$$π$$2$$ω$$t2=$$π$$2$$ or $$t=$$πω

If vector $\vec{A} = cosωt \hat{i} + sinωt \hat{j}$ and $\vec{B} = \cos \frac{ωt}{2} \hat{i} + \sin \frac{ωt}{2} \hat{j}$ are functions of time, then the value of t at which they are orthogonal to each other is

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If vector A→=cosωti^+sinωtj^ and B→=cosωt2i^+sinωt2j^ are functions of time, then the value of t at which they are orthogonal to each other is

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If vector $\vec{A} = cosωt \hat{i} + sinωt \hat{j}$ and $\vec{B} = \cos \frac{ωt}{2} \hat{i} + \sin \frac{ωt}{2} \hat{j}$ are functions of time, then the value of t at which they are orthogonal to each other is