Vectors 5 Question 5
5. A non-zero vector $\overrightarrow{\mathbf{a}}$ is parallel to the line of intersection
of the plane determined by the vectors $\hat{\mathbf{i}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}$ and the plane determined by the vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$. The angle between $\overrightarrow{\mathbf{a}}$ and the vector $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is…… .
(1996, 2M)
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Answer:
Correct Answer: 5. $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
Solution:
- Equation of the plane containing $\hat{\mathbf{i}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ is
$$ \begin{aligned} & {[(\overrightarrow{\mathbf{r}}-\hat{\mathbf{i}}) \quad \hat{\mathbf{i}} \quad(\hat{\mathbf{i}}+\hat{\mathbf{j}})]=0} \\ & \Rightarrow \quad(\overrightarrow{\mathbf{r}}-\hat{\mathbf{i}}) \cdot[\hat{\mathbf{i}} \times(\hat{\mathbf{i}}+\hat{\mathbf{j}})]=0 \\ & \Rightarrow \quad{(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-\hat{\mathbf{i}}} \cdot[\hat{\mathbf{i}} \times \hat{\mathbf{i}}+\hat{\mathbf{i}} \times \hat{\mathbf{j}}]=0 \\ & \Rightarrow \quad{(x-1) \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}} \cdot[\hat{\mathbf{k}}]=0 \\ & \Rightarrow \quad(x-1) \hat{\mathbf{i}} \cdot \hat{\mathbf{k}}+y \hat{\mathbf{j}} \cdot \hat{\mathbf{k}}+z \hat{\mathbf{k}} \cdot \hat{\mathbf{k}}=0 \\ & \Rightarrow \quad z=0 \end{aligned} $$
Equation of the plane containing $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{k}}$ is
$$ \begin{aligned} & {[(\overrightarrow{\mathbf{r}}-(\hat{\mathbf{i}}-\hat{\mathbf{j}}))(\hat{\mathbf{i}}-\hat{\mathbf{j}})(\hat{\mathbf{i}}+\hat{\mathbf{k}})]=0} \\ & \Rightarrow \quad(\overrightarrow{\mathbf{r}}-\hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot[(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}+\hat{\mathbf{k}})]=0 \\ & \Rightarrow{(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(\hat{\mathbf{i}}-\hat{\mathbf{j}})} \cdot[\hat{\mathbf{i}} \times \hat{\mathbf{i}}+\hat{\mathbf{i}} \times \hat{\mathbf{k}}-\hat{\mathbf{j}} \times \hat{\mathbf{i}}-\hat{\mathbf{j}} \times \hat{\mathbf{k}}]=0 \\ & \Rightarrow \quad{(x-1) \hat{\mathbf{i}}+(y+1) \hat{\mathbf{j}}+z \hat{\mathbf{k}})} \cdot[-\hat{\mathbf{j}}+\hat{\mathbf{k}}-\hat{\mathbf{i}}]=0 \\ & \Rightarrow \quad-(x-1)-(y+1)+z=0 \end{aligned} $$
Let $\overrightarrow{\mathbf{a}}=a _1 \hat{\mathbf{i}}+a _2 \hat{\mathbf{j}}+a _3 \hat{\mathbf{k}}$
Since, $\overrightarrow{\mathbf{a}}$ is parallel to Eqs. (i) and (ii), we obtain
$$ a _3=0 $$
and $a _1+a _2-a _3=0 \Rightarrow a _1=-a _2, a _3=0$
Thus, a vector in the direction $\overrightarrow{\mathbf{a}}$ is $\hat{\mathbf{i}}-\hat{\mathbf{j}}$.
If $\theta$ is the angle between $\overrightarrow{\mathbf{a}}$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$, then
$$ \begin{aligned} \cos \theta & = \pm \frac{(1)(1)+(-1)(-2)}{\sqrt{1+1} \sqrt{1+4+4}}= \pm \frac{3}{\sqrt{2} \cdot 3} \\ \Rightarrow \quad \cos \theta & = \pm \frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4} \text { or } \frac{3 \pi}{4} \end{aligned} $$
