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) ${\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}} _2, \overrightarrow{\mathbf{c}} _4}$
Show Answer
Answer:
Correct Answer: 12. $\overrightarrow{\mathbf{v}} _1=2 \hat{\mathbf{i}}, \overrightarrow{\mathbf{v}} _2=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\overrightarrow{\mathbf{v}} _3=3 \hat{\mathbf{i}} \pm 2 \hat{\mathbf{j}} \pm 4 \hat{\mathbf{k}}$
Solution:
- 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.
