`vec(a)+vec(b)+vec(c)=vec(0)` such that `|vec(a)|=3, |vec(b)|=5 and |vec(c)|=7`.
What is `vec (a). vec(b) + vec(b). vec(c) + vec(c). vec (a)` equal to ?

Options

A. -83
B. `-83//2`
C. 75
D. `-75//2`

Answer

B

Explanation

We have, `vec(a)+vec(b)+vec(c)=0`
On squaring both sides request.
`vec(a)^(2)+vec(b)^(2)+vec(c)^(2)+2vec(a)vec(b)+2vec(b)vec(c)+2vec(c)vec(a)=0 (. :' vec(a).vec(b)=vec(b).vec(a),vec(b).vec(c)=vec(c).vec(b) and vec(c).vec(a)=vec(a).vec(c))`
`rArr |vec(a)|^(2)+|vec(b)|^(2)+|vec(c)|^(2)=-2[a.b+b.c+c.a]`
`rArr (3)^(2)+(5)^(2)+(7)^(2)=-2[a.b+b.c+c.a]`
`rArr a.b+b.c+c.a=(9+25)+49)/(-2)=-(83)/(2)`
Now a+b+c+=0 [usingeq. (i)]
`rArr a+b=-c`
On squaring both sides, we get
`rArr a^(2)+b^(2)+2a.b=c^(2)`
`rArr (3)^(2)+(5)^(2)+2ab=(7)^(2)`
`rArr vec(a).vec(b)=(15)/(2)`
`rArr |vec(a)||vec(b)|cos theta = (15)/(2)rArr 3.5 cos theta = (15)/(2)`
`rArr cos theta = (1)/(2) = cos. (pi)/(3)`
`:. theta =(pi)/(3)`
From eq. (i),
`vec(b)+vec(c)=-vec(a)`
`rArr b^(2)+c^(2)+2b.c=a^(2)`
`rArr 2b.c=a^(2)-b^(2)-c^(2)=9-25-49=-65`
`rArr b.c=-(65)/(2) rArr |b||c|cos theta = -(65)/(2)`
`rArr cos theta = (65)/(2)xx(1)/(5)xx(1)/(7)=-(13)/(14)`
Also, `vec(a)+vec(b)|=|-vec(c)|=|vec(c)|=7`