Vectors 1 Question 24
25. Find 3-dimensional vectors $\overrightarrow{\mathbf{v}} _1, \overrightarrow{\mathbf{v}} _2, \overrightarrow{\mathbf{v}} _3$ satisfying
$$ \begin{aligned} & \overrightarrow{\mathbf{v} _1} \cdot \overrightarrow{\mathbf{v}} _1=4, \overrightarrow{\mathbf{v} _1} \cdot \overrightarrow{\mathbf{v}} _2=-2, \overrightarrow{\mathbf{v} _1} \cdot \overrightarrow{\mathbf{v}} _3=6 \\ & \overrightarrow{\mathbf{v}} _2 \cdot \overrightarrow{\mathbf{v}} _2=2, \overrightarrow{\mathbf{v}} _2 \cdot \overrightarrow{\mathbf{v}} _3=-5, \overrightarrow{\mathbf{v}} _3 \cdot \overrightarrow{\mathbf{v}} _3=29 \end{aligned} $$
(2001, 5M)
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Answer:
Correct Answer: 25. (b)
Solution:
- We have, $\left|\overrightarrow{\mathbf{v}} _1\right|=2,\left|\overrightarrow{\mathbf{v}} _2\right|=\sqrt{2}$ and $\left|\overrightarrow{\mathbf{v}} _3\right|=\sqrt{29}$
If $\theta$ is the angle between $\overrightarrow{\mathbf{v}} _1$ and $\overrightarrow{\mathbf{v}} _2$, then
$$ \begin{array}{rlrl} 2 \sqrt{2} \cos \theta & =-2 \\ \Rightarrow \quad \cos \theta & =-\frac{1}{\sqrt{2}} \\ \Rightarrow \quad & \theta & =135^{\circ} \end{array} $$
Since, any two vectors are always coplanar and data is not sufficient, so we can assume $\overrightarrow{\mathbf{v}} _1$ and $\overrightarrow{\mathbf{v}} _2$ in $x-y$ plane.
$$ \begin{aligned} & \overrightarrow{\mathbf{v}} _1=2 \hat{\mathbf{i}} \\ & \overrightarrow{\mathbf{v}} _2=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \end{aligned} $$
and
$$ \overrightarrow{\mathbf{v} _3}=\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}} $$
Since, $\quad \overrightarrow{\mathbf{v}} _3 \cdot \overrightarrow{\mathbf{v}} _1=6=2 \alpha \quad \Rightarrow \quad \alpha=3$
Also, $\quad \overrightarrow{\mathbf{v}} _3 \cdot \overrightarrow{\mathbf{v}} _2=-5=-\alpha \pm \beta \Rightarrow \beta= \pm 2$
and $\quad \overrightarrow{\mathbf{v}} _3 \cdot \overrightarrow{\mathbf{v}} _3=29=\alpha^{2}+\beta^{2}+\gamma^{2} \Rightarrow \gamma= \pm 4$
Hence, $\quad \overrightarrow{\mathbf{v}} _3=3 \hat{\mathbf{i}} \pm 2 \hat{\mathbf{j}} \pm 4 \hat{\mathbf{k}}$
