Question

Find the vector equation of the plane passing through three points with position vectors $\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,$, $2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$ and $\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$. Also find the coordinates of the point of intersection of this plane and the line $\overrightarrow{r}=3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\,+\lambda \left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)$.

Answer 1

Hint: We start solving this problem by considering the given position vectors as $A,B,C$and then we find $\overrightarrow{AB}$ and $\overrightarrow{BC}$ using the formula $\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$. Then we find the vector that is normal to the plane passing through $A,B,C$ using the formula $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{BC}$. Then we find the equation of the required plane passing through any one of the given points and normal to the obtained vector using the formula $\left( \overrightarrow{r}-\overrightarrow{a} \right).\overrightarrow{n}=0$. Then we solve the obtained vector equation and the given equation of line to get the value of $\lambda $. Hence, we get the coordinates of the required point.

Complete step by step answer:
Let us consider $A=\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,$, $B=2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$and $C=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$.
Now, let us consider the formula for line joining A and B, $\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$
By applying the above formula, we get,
$\begin{align}
  & \Rightarrow \overrightarrow{AB}=\left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)-\left( \overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\, \right) \\
 & \Rightarrow \overrightarrow{AB}=\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \\
 & \Rightarrow \overrightarrow{BC}=\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)-\left( 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right) \\
 & \Rightarrow \overrightarrow{BC}=-\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{j}}\, \\
\end{align}$

Now, we find the vector normal to the plane passing through the points $\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,$, $2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$ and $\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$.
Let us consider the formula, vector normal to the plane passing through the points $A,B,C$ is $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$
By applying the above formula, we get,
$\begin{align}
  & \overrightarrow{n}=\left| \begin{matrix}
   \overset{\wedge }{\mathop{i}}\, & \overset{\wedge }{\mathop{j}}\, & \overset{\wedge }{\mathop{k}}\, \\
   1 & -2 & 3 \\
   -1 & 3 & 0 \\
\end{matrix} \right| \\
 & \overrightarrow{n}=\overset{\wedge }{\mathop{i}}\,\left( -2\left( 0 \right)-3\left( 3 \right) \right)-\overset{\wedge }{\mathop{j}}\,\left( 1\left( 0 \right)-3\left( -1 \right) \right)+\overset{\wedge }{\mathop{k}}\,\left( 1\left( 3 \right)-\left( -2 \right)\left( -1 \right) \right) \\
 & \overrightarrow{n}=-9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \\
\end{align}$
Now, let us consider another formula, that is,
The equation of the plane passing through the point $\overrightarrow{a}$ and is normal to the vector $\overrightarrow{n}$ is given by $\left( \overrightarrow{r}-\overrightarrow{a} \right).\overrightarrow{n}=0$
Now, by using the above formula, the equation of the plane passing through the point with position vector $\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,$ and normal to the vector $-9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,$is given by
$\begin{align}
  & \left( \overrightarrow{r}-\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\, \right).\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=0 \\
 & \Rightarrow \overrightarrow{r}.\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)-\left( \overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\, \right).\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=0 \\
 & \Rightarrow \overrightarrow{r}.\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=\left( \overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\, \right).\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right) \\
 & \Rightarrow \overrightarrow{r}.\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=-9-3-2 \\
 & \Rightarrow \overrightarrow{r}.\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=-14...............\left( 1 \right) \\
\end{align}$
Now, consider the given equation of the line, $\overrightarrow{r}=3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\,+\lambda \left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)...............\left( 2 \right)$
By solving the equations (1) and (2), we get,
$\begin{align}
  & \left( 3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\,+\lambda \left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right) \right).\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=-14 \\
 & \Rightarrow \left( \left( 3+2\lambda \right)\overset{\wedge }{\mathop{i}}\,+\left( -1-2\lambda \right)\overset{\wedge }{\mathop{j}}\,+\left( -1+\lambda \right)\overset{\wedge }{\mathop{k}}\, \right).\left( -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right)=-14 \\
 & \Rightarrow -27-18\lambda +3+6\lambda -1+\lambda =-14 \\
 & \Rightarrow -11-11\lambda =0 \\
 & \Rightarrow -11\lambda =11 \\
 & \Rightarrow \lambda =-\dfrac{11}{11} \\
 & \Rightarrow \lambda =-1 \\
\end{align}$
Now, by substituting the value of $\lambda $in equation (2), we get,
$\begin{align}
  & \overrightarrow{r}=3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\,+\left( -1 \right)\left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, \right) \\
 & \Rightarrow \overrightarrow{r}=3\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\,-2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,-\overset{\wedge }{\mathop{k}}\, \\
 & \Rightarrow \overrightarrow{r}=\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\, \\
\end{align}$

Hence, the coordinates of the required point are $\left( 1,1,-2 \right)$.

Note: There is a possibility of making a mistake by considering $\overrightarrow{AB}=\overrightarrow{A}-\overrightarrow{B}$ instead of $\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$, when finding the vector joining two vectors we should subtract the second one from the first. One might also make a mistake by considering $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{BC}$ instead of $\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$ while solving the question. So, one should take care of which vector they are using while solving the problem.