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Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. A vector coplanar to $\mathbf{a}$ and $\mathbf{b}$ has a projection along c of magnitude $\frac{1}{\sqrt{3}}$, then the vector is
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The correct answer is:
$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
Let vector $\mathbf{r}$ be coplanar to $\mathbf{a}$ and $\mathbf{b}$.
$$
\begin{array}{ll}
\therefore & \mathbf{r}=\mathbf{a}+t \mathbf{b} \\
\Rightarrow & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=\hat{\mathbf{i}}(1+t)+\hat{\mathbf{j}}(2-t)+\hat{\mathbf{k}}(1+t)
\end{array}
$$
The projection of $\mathbf{r}$ on $\mathbf{c}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ $\Rightarrow|2-t|=\pm 1 \Rightarrow t=1$ or 3
When, $t=1$ we have $\mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
When, $t=3$ we have $\mathbf{r}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
$$
\begin{array}{ll}
\therefore & \mathbf{r}=\mathbf{a}+t \mathbf{b} \\
\Rightarrow & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=\hat{\mathbf{i}}(1+t)+\hat{\mathbf{j}}(2-t)+\hat{\mathbf{k}}(1+t)
\end{array}
$$
The projection of $\mathbf{r}$ on $\mathbf{c}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ $\Rightarrow|2-t|=\pm 1 \Rightarrow t=1$ or 3
When, $t=1$ we have $\mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
When, $t=3$ we have $\mathbf{r}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
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