Vectors 1 Question 11

12. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are three non-zero, non-coplanar vectors and $\overrightarrow{\mathbf{b}} _1=\overrightarrow{\mathbf{b}}-\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b} _2}=\overrightarrow{\mathbf{b}}+\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}$ $\overrightarrow{\mathbf{c} _1}=\overrightarrow{\mathbf{c}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|^{2}} \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c} _2}=\overrightarrow{\mathbf{c}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b} _1}}{|\overrightarrow{\mathbf{b}}|^{2}} \overrightarrow{\mathbf{b} _1}$, $\overrightarrow{\mathbf{c} _3}=\overrightarrow{\mathbf{c}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}} _2}{\left|\overrightarrow{\mathbf{b}} _2\right|^{2}} \overrightarrow{\mathbf{b} _2}, \overrightarrow{\mathbf{c} _4}=\overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}$

Then, which of the following is a set of mutually orthogonal vectors?

(2005, 1M)

(a) ${\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}} _1, \overrightarrow{\mathbf{c}} _1}$

(b) ${\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b} _1}, \overrightarrow{\mathbf{c} _2}}$

(c) ${\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b} _2}, \overrightarrow{\mathbf{a} _3}}$

(d) ${\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}} _2, \overrightarrow{\mathbf{c}} _4}$

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Answer:

Correct Answer: 12. (b)

Solution:

  1. Since, $\overrightarrow{\mathbf{b} _1}=\overrightarrow{\mathbf{b}}-\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b} _1}=\overrightarrow{\mathbf{b}}+\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}$

and $\quad \overrightarrow{\mathbf{c} _1}=\overrightarrow{\mathbf{c}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|^{2}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c} _2}=\overrightarrow{\mathbf{c}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b} _1}}{|\overrightarrow{\mathbf{b}}|^{2}} \overrightarrow{\mathbf{b} _1}$

$ \overrightarrow{\mathbf{c} _3}=\overrightarrow{\mathbf{c}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b} _2}}{\left|\overrightarrow{\mathbf{b}} _2\right|^{2}} \overrightarrow{\mathbf{b} _2}, \overrightarrow{\mathbf{c} _4}=\overrightarrow{\mathbf{a}}-\frac{\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^{2}} \overrightarrow{\mathbf{a}} $

which shows $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b} _1}=0=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c} _2}=\overrightarrow{\mathbf{b} _1} \cdot \overrightarrow{\mathbf{c} _2}$

So, ${\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}} _1, \overrightarrow{\mathbf{c}} _2}$ are mutually orthogonal vectors.



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