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Three vectors `vecA,vecB` and `vecC` are such that `vecA=vecB+vecC` and their magnitudes are in ratio 5:4:3 respectively. Find angle between vector `vecA` and `vecC` |
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Answer» (a) Given that : `vecA=vecB+vecCrArrvecA-vecC=vecB` By taking self dot product on both sides `(vecA-vecC).(vecA-vecC)=vecB.vecBrArrA^(2)+C^(2)-2vecA.vecC=B^(2)` Now let angle between `vecA` and `vecC` be `theta` then `A^(2)+C^(2)-2AC cos theta=B^(2)` `theta cos theta=(A^(2)+C^^(2)-B^(2))/(2AC)=((5)^(2)+(3)^(2)-(4)^(2))/(2(5)(3))=(18)/(30)=(3)/(5)rArr=cos^(-1)((3)/(5))=53^(@)` or Since `5^(2)=4^(2)+3^(2)` the vectors `vecA, vecB` and `vecC` with `vecA=vecB+vecC` ,ale a troamg,e wotj amg,e netweem `vecB` and `vecC` as `90^(@)`. If `theta` is the angle between `vecA` and `vecC`, then `cos theta =(3)/(5)=0.6=cos (53)^(@) therefore theta=53^(@)` |
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