Vectors 1 Question 18
19. Let $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ be three vectors. A vector in the plane of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$, whose projection on $\overrightarrow{\mathbf{a}}$ is of magnitude $\sqrt{2 / 3}$, is
(1993, 2M)
(a) $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
(b) $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
(c) $-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
(d) $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
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Answer:
Correct Answer: 19. (3)
Solution:
- Given vectors are $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. Any vector $\overrightarrow{\mathbf{r}}$ in the plane of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ is $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{b}}+t(\overrightarrow{\mathbf{c}})=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}+t(\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$
$$ =(1+t) \hat{\mathbf{i}}+(2+t) \hat{\mathbf{j}}-(1+2 t) \hat{\mathbf{k}} $$
Since, projection of $\overrightarrow{\mathbf{r}}$ on $\overrightarrow{\mathbf{a}}$ is $\sqrt{\frac{2}{3}}$.
$$ \therefore \quad \frac{\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\sqrt{\frac{2}{3}} $$
$$ \begin{aligned} & \Rightarrow\left|\frac{2(1+t)-(2+t)-(1+2 t)}{\sqrt{6}}\right|=\sqrt{\frac{2}{3}} \\ & \Rightarrow \quad|-(1+t)|=2 \Rightarrow t=1 \text { or }-3 \end{aligned} $$
On putting $t=1,-3$ in Eq. (i) respectively, we get
$$ \begin{aligned} & \overrightarrow{\mathbf{r}} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ \text { or } & \overrightarrow{\mathbf{r}} & =-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}} \end{aligned} $$
