Vectors 1 Question 14
15. Let $\overrightarrow{\mathbf{u}}, \overrightarrow{\mathbf{v}}$ and $\overrightarrow{\mathbf{w}}$ be vectors such that $\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}=\overrightarrow{\mathbf{0}}$. If $|\overrightarrow{\mathbf{u}}|=3,|\overrightarrow{\mathbf{v}}|=4$ and $|\overrightarrow{\mathbf{w}}|=5$, then $\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}}$ is
(1995, 2M)
(a) 47
(b) -25
(c) 0
(d) 25
Show Answer
Answer:
Correct Answer: 15. $(3: 2)$
Solution:
- Since, $\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}=\overrightarrow{\mathbf{0}} \Rightarrow|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}|^{2}=0$
$\Rightarrow|\overrightarrow{\mathbf{u}}|^{2}+|\overrightarrow{\mathbf{v}}|^{2}+|\overrightarrow{\mathbf{w}}|^{2}+2(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}})=0$
$\Rightarrow \quad 9+16+25+2(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}})=0$
$\Rightarrow \quad \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}}=-25$
