Vectors 1 Question 10
11. Let, $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$.A vector coplanar to $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ has a projection along $\overrightarrow{\mathbf{c}}$ of magnitude $\frac{1}{\sqrt{3}}$, then the vector is
(2006, 3M)
(a) $4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
(b) $4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
(c) $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
(d) None of these
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Answer:
Correct Answer: 11. (a)
Solution:
- Let vector $\overrightarrow{\mathbf{r}}$ be coplanar to $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$.
$ \begin{array}{ll} \therefore & \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+t \overrightarrow{\mathbf{b}} \\ \Rightarrow & \overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{array} $
$ =\hat{\mathbf{i}}(1+t)+\hat{\mathbf{j}}(2-t)+\hat{\mathbf{k}}(1+t) $
The projection of $\overrightarrow{\mathbf{r}}$ on $\overrightarrow{\mathbf{c}}=\frac{1}{\sqrt{3}}$.
[given]
$ \begin{aligned} & \Rightarrow \quad \frac{\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{c}}}{|\overrightarrow{\mathbf{c}}|}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow(2-t)= \pm 1 \Rightarrow t=1 \text { or } 3 \end{aligned} $
When, $t=1$, we have $\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
When, $t=3$, we have $\overrightarrow{\mathbf{r}}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
