Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Any vector in the plane of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ is $\overrightarrow{\mathbf{r}}=m\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}$ $$=(m+1)\hat{\mathbf{i}}+(2m+1)\hat{\mathbf{j}}+(-m-2)\hat{\mathbf{k}}$$ Projection of $\overrightarrow{\mathbf{r}}$ on $\overrightarrow{\mathbf{a}}=\frac{\overrightarrow{\mathbf{r}}\cdot \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\left|\sqrt{\frac{2}{3}}\right|$ $$\begin{array}{l}\therefore \frac{2(m+1)-(2m+1)+(-m-2)}{\sqrt{6}}=\pm \sqrt{\frac{2}{3}}\\ \Rightarrow -m-1=\pm 2\end{array}$$ $\Rightarrow$ $m=-3$ and 1 Hence, $\overrightarrow{\mathbf{r}}=-2\hat{\mathbf{i}}-5\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{r}}=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}-3\hat{\mathbf{k}}$