Question

Find the vector equation of the plane that contains the lines $\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right)$ and the point (-1, 3, -4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained.

Answer 1

Given line $\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\lambda \left(\hat{i}+2\hat{j}-\hat{k}\right)$ passes through (1,1,0) and is parallel to the vector $\hat{i}+2\hat{j}-\hat{k}$

i.e. The given plane passes through (1,1,0) and B(−1, 3, −4) and is parallel to $\overrightarrow{b}=\overrightarrow{c}+2\hat{j}-\hat{k}$

Let $\overrightarrow{n}$ be the normal vector to the required plane.

Then $\overrightarrow{n}$ is perpendicular to both $\overrightarrow{b}\mathrm{and}\overrightarrow{AB}$

i.e. $\overrightarrow{n}$ is parallel to $\overrightarrow{AB}\times \overrightarrow{B}$
$$\begin{gathered} \mathrm{Let}{\overrightarrow{n}}_1=\overrightarrow{AB}\times \overrightarrow{b} \\ \mathrm{then}{\overrightarrow{n}}_1=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 2& -4\\ 1& 2& -1\end{array}\right| \\ =\hat{i}\left(-2+8\right)-\hat{j}\left(2+4\right)+R\left(-4-2\right) \\ {\overrightarrow{n}}_1=6\hat{i}-6\hat{j}-6\hat{k} \\ \mathrm{Let}\overrightarrow{\alpha }\mathrm{be}\mathrm{position}\mathrm{vector}\mathrm{of}A\mathrm{i}.\mathrm{e}.\overrightarrow{\alpha }=\hat{i}+\hat{j} \end{gathered}$$

Then the required plane passes through $\overrightarrow{\alpha }=\hat{i}+\hat{j}$ and is perpendicular to ${\overrightarrow{n}}_1=6\hat{i}-6\hat{j}-6\hat{k}$
So, its vector equation is
$$\begin{gathered} \left(\overrightarrow{r}-\overrightarrow{\alpha }\right).{\overrightarrow{n}}_1=0 \\ \mathrm{i}.\mathrm{e}.\overrightarrow{r}.{\overrightarrow{n}}_1=\overrightarrow{\alpha }.{\overrightarrow{n}}_1 \\ \mathrm{i}.\mathrm{e}.\overrightarrow{r}.\left(6\hat{i}-6\hat{j}-6\hat{k}\right)=\left(\hat{i}+\hat{j}\right).\left(6\hat{i}-6\hat{j}-6\hat{k}\right) \\ \overrightarrow{r}.\left(6\hat{i}-6\hat{j}-6\hat{k}\right)=6-6=0 \\ \mathrm{i}.\mathrm{e}.\overrightarrow{r}.\left(\hat{i}-\hat{j}-\hat{k}\right)=0 \\ \mathrm{for}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k} \end{gathered}$$
Equation of plane is x – y – z = 0
The length of perpendicular form (2, 1, 4) to plane x – y – z = 0 is
$$\begin{gathered} d=\frac{\left|\left(2\hat{i}+\hat{j}+4\hat{k}\right).\left(\hat{i}-\hat{j}-\hat{k}\right)\right|}{\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2}}=\frac{\left|2-1-4\right|}{\sqrt{3}} \\ =\frac{3}{\sqrt{3}}=\sqrt{3} \\ \mathrm{i}.\mathrm{e}.d=\sqrt{3} \end{gathered}$$