Vectors 1 Question 22
23. Let $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}, \overrightarrow{\mathbf{C}}$ be vectors of length $3,4,5$ respectively. Let $\overrightarrow{\mathbf{A}}$ be perpendicular to $\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}, \overrightarrow{\mathbf{B}}$ to $\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{C}}$ to $\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}$. Then, the length of vector $\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}$ is … .
(1981, 2M)
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Answer:
Correct Answer: 23. (b)
Solution:
- Given, $\quad|\overrightarrow{\mathbf{A}}|=3,|\overrightarrow{\mathbf{B}}|=4,|\overrightarrow{\mathbf{C}}|=5$
Since, $\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}})=\overrightarrow{\mathbf{B}} \cdot(\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{A}})=\overrightarrow{\mathbf{C}} \cdot(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}})=0$
$\therefore|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}|^{2}=|\overrightarrow{\mathbf{A}}|+|\overrightarrow{\mathbf{B}}|^{2}+|\overrightarrow{\mathbf{C}}|^{2}$
$$ +2(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{B}} \cdot \overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{C}} \cdot \overrightarrow{\mathbf{A}}) $$
$$ =9+16+25+0 $$
[from Eq. (i), $\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{B}} \cdot \overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{C}} \cdot \overrightarrow{\mathbf{A}}=0$ ]
$$ \begin{array}{ll} \therefore & |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}|^{2}=50 \\ \Rightarrow & |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}|=5 \sqrt{2} \end{array} $$
