Three vectors a→, b→ and c→ satisfy a→+b→+c→=0→. If a→=3,b→=4, c→=2, then a→.b→+b→·c→+c→.a→+2(|a→|+|b→|+|c→|)=

Correct answer: (B) 7/2. Given, a→+b→+c→=0 ⇒a→+b→+c→^2=0 ⇒a→^2+b→^2+c→^2+2a→.b→+b→.c→+c→.a→=0 ⇒9+16+4+2a→.b→+b→.c→+c→.a→=0 ⇒a→.b→+b→.c→+c→.a→=-29/2 ∴a→.b→+b→·c→+c→.a→+2(|a→|+|b→|+|c→|)=-29/2+18=7/2

Q73 / 160 · TS EAMCET 2021 (04 Aug Shift 1) · Mathematics

Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} + \vec{b} + \vec{c} = \vec{0} .$ If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 2,$ then $\vec{a} . \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} . \vec{a} + 2 (| \vec{a} | + | \vec{b} | + | \vec{c} |) =$

A
$\frac{- 7}{2}$
Correct answer
B
$\frac{7}{2}$
C
$\frac{- 11}{2}$
D
$\frac{11}{2}$

Solution

Given, $\vec{a} + \vec{b} + \vec{c} = 0$

$\Rightarrow {|\vec{a} + \vec{b} + \vec{c}|}^{2} = 0$

$\Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0$

$\Rightarrow 9 + 16 + 4 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0$

$\Rightarrow \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} = - \frac{29}{2}$

$∴ \vec{a} . \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} . \vec{a} + 2 (| \vec{a} | + | \vec{b} | + | \vec{c} |) = - \frac{29}{2} + 18 = \frac{7}{2}$

Hence, the correct answer is option: 'B'.

Three vectors a→, b→ and c→ satisfy a→+b→+c→=0→. If a→=3,b→=4, c→=2, then a→.b→+b→·c→+c→.a→+2(|a→|+|b→|+|c→|)=

Solution

Relevant Questions

Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} + \vec{b} + \vec{c} = \vec{0} .$ If $|\vec{a}| = 3, |\vec{b}| = 4, |\vec{c}| = 2,$ then $\vec{a} . \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} . \vec{a} + 2 (| \vec{a} | + | \vec{b} | + | \vec{c} |) =$