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

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Answer:

Correct Answer: 15. (b)

Solution:

  1. Since, $\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}=\overrightarrow{\mathbf{0}} \Rightarrow|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}|^{2}=0$

$\Rightarrow|\overrightarrow{\mathbf{u}}|^{2}+|\overrightarrow{\mathbf{v}}|^{2}+|\overrightarrow{\mathbf{w}}|^{2}+2(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}})=0$

$\Rightarrow \quad 9+16+25+2(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}})=0$

$\Rightarrow \quad \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{u}}=-25$



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