Question

force acting on a particle get to ICAP + 3 J cap - theta Newton produces a displacement of ICAP + 3 J cap - 2 J cap metre then find the work done​

Answer 1

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Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>a-2b+3c=0

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>a-2b+3c=0a+2b-c=0

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>a-2b+3c=0a+2b-c=0=> a/(2-6)=b/(3+1)=c/(2+2)

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>a-2b+3c=0a+2b-c=0=> a/(2-6)=b/(3+1)=c/(2+2)=>a/(-1)=b/1=c/1

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>a-2b+3c=0a+2b-c=0=> a/(2-6)=b/(3+1)=c/(2+2)=>a/(-1)=b/1=c/1Thus A=-i+j+k.

Let A=ai+bj+ck be a vector which is perpendicular to B=i-2j+3k and C=i+2j-k.A.B=0 and A.C=0=>a-2b+3c=0a+2b-c=0=> a/(2-6)=b/(3+1)=c/(2+2)=>a/(-1)=b/1=c/1Thus A=-i+j+k.Hence required unit vector along vector A=(-i+j+k)/√(3).

Answer 2

It can be find by the help of dot product.

Solution is in attachment.

Thank You