Mathematics › Vectors › Product of 2 vectors
If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \quad \overline{\mathrm{b}}=2…
If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \quad \overline{\mathrm{b}}=2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \quad$ and $\quad \overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}}$, $\overline{\mathrm{r}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}$, then the value $\frac{\overline{\mathrm{r}}}{|\overline{\mathrm{r}}|}$ is
$\frac{\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{11}}$ $\frac{\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{11}}$ $\frac{\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{11}}$ $\frac{\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{11}}$ Solution
Let $\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$ then
$\begin{aligned}
& \overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \quad \Rightarrow(\overline{\mathrm{r}}-\overline{\mathrm{b}}) \times \overline{\mathrm{a}}=\overline{0} \\
& \therefore \quad\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
x & y-2 & \mathrm{z}+1 \\
1 & 1 & 0
\end{array}\right|=\overline{0} \\
& \Rightarrow(-\mathrm{z}-1) \hat{\mathrm{i}}-(-\mathrm{z}-1) \hat{\mathrm{j}}+(x-y+2) \hat{\mathrm{k}}=\overline{0} \\
& \Rightarrow \mathrm{z}=-1, x-y=-2
\end{aligned}$
Now, $\overline{\mathrm{r}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}=(\overline{\mathrm{r}}-\overline{\mathrm{a}}) \times \overline{\mathrm{b}}=\overline{0}$
$\begin{aligned}
\therefore \quad & \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
x-1 & y-1 & z \\
0 & 2 & -1
\end{array}\right|=\overline{0} \\
& \Rightarrow(1-y-2 \mathrm{z}) \hat{\mathrm{i}}-(1-x) \hat{\mathrm{j}}+(2 x-2) \hat{\mathrm{k}}=\overline{0} \\
& \Rightarrow 1-y-2 z=0, x=1
\end{aligned}$
Solving (i) and (ii), we get
$\begin{aligned}
& x=1, y=3, z=-1 \\
\therefore \quad & \bar{r}=\hat{i}+3 \hat{j}-\hat{k} \\
& |\bar{r}|=\sqrt{1+9+1}=\sqrt{11} \\
\therefore \quad & \frac{\bar{r}}{\left|-\frac{\mathrm{r}}{-}\right|}=\frac{\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{11}}
\end{aligned}$
Asked in: MHT CET 2023 (11 May Shift 1)
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