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Find the angle between the vectors hat(i)-2hat(j)+3hat(k) and 3hat(i)-2hat(j)+hat(k). |
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Answer» Solution :Let `vec(a)=HAT(i)-2hat(j)+3hat(k) and vec(b)=3hat(i)-2hat(j)+hat(k)`. Magnitude of `vec(a),|vec(a)|=sqrt(1^2+(-2)^2+3^2)=sqrt(1+4+9)=sqrt(14)`. Magnitude of `vec(b),|vec(b)|=sqrt(3^2+(-2)^2+1^2)=sqrt(9+4+1)=sqrt(14)`. Now, `vec(a)vec(b)=(hat(i)-2hat(j)+3hat(k)).(3hat(i)-2hat(j)+hat(k))=1.3+(-2).(-2)+3.1=3+4+3=10` (Do product of TWO VECTORS is equal to the sum of the products of their corresponding components.) Let `theta` be the required angle between a and b then `cos theta =(vec(a).vec(b))/(|vec(a)||vec(b)|)=(10)/(sqrt(14)sqrt(14))=(10)/(14)=(5)/(7)rArr theta= cos^(-1)((5)/(7))`. |
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