1.

If vecA=3hati+4hatj and vecB=7hati+24hatj find a vector having thesame magnitude as vecB and parallel to vecA.

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Solution :Now `vecAxxvecB`=(hati+2hatj+3hatk)XX(2hati-hatj+hatk)=-hatk-hatj-4hatk+2hatj+6hatj+3hati=5hati+5hatj-5hatk`
`abs(vecAxxvecB)=Absintheta` we get `SIN theta = abs(vecAxxvecB)//AB`
SINCE A=`sqrt(1+4+9)=SQRT14` and B=sqrt(4+1+1)=sqrt6`


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