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Question: Find the vector equation of the plane passing through three points with position vectors \(\overset{...

Find the vector equation of the plane passing through three points with position vectors i∧ +j∧ −2k∧ \overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,, 2i∧ −j∧ +k∧ 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, and i∧ +2j∧ +k∧ \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 r→=3i∧ −j∧ −k∧ +λ(2i∧ −2j∧ +k∧ )\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).

Explanation

Solution

We start solving this problem by considering the given position vectors as A,B,CA,B,Cand then we find AB→\overrightarrow{AB} and BC→\overrightarrow{BC} using the formula AB→=B→−A→\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}. Then we find the vector that is normal to the plane passing through A,B,CA,B,C using the formula n→=AB→×BC→\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 (r→−a→).n→=0\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 λ\lambda . Hence, we get the coordinates of the required point.

Complete step by step answer:
Let us consider A=i∧ +j∧ −2k∧ A=\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,, B=2i∧ −j∧ +k∧ B=2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,and C=i∧ +2j∧ +k∧ 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, AB→=B→−A→\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}
By applying the above formula, we get,
⇒AB→=(2i∧ −j∧ +k∧ )−(i∧ +j∧ −2k∧ ) ⇒AB→=i∧ −2j∧ +3k∧  ⇒BC→=(i∧ +2j∧ +k∧ )−(2i∧ −j∧ +k∧ ) ⇒BC→=−i∧ +3j∧  \begin{aligned} & \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{aligned}

Now, we find the vector normal to the plane passing through the points i∧ +j∧ −2k∧ \overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\,, 2i∧ −j∧ +k∧ 2\overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\, and i∧ +2j∧ +k∧ \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,CA,B,C is n→=AB→×AC→\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}
By applying the above formula, we get,
n→=∣i∧ j∧ k∧  1−23 −130 ∣ n→=i∧ (−2(0)−3(3))−j∧ (1(0)−3(−1))+k∧ (1(3)−(−2)(−1)) n→=−9i∧ −3j∧ +k∧  \begin{aligned} & \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{aligned}
Now, let us consider another formula, that is,
The equation of the plane passing through the point a→\overrightarrow{a} and is normal to the vector n→\overrightarrow{n} is given by (r→−a→).n→=0\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 i∧ +j∧ −2k∧ \overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,-2\overset{\wedge }{\mathop{k}}\, and normal to the vector −9i∧ −3j∧ +k∧ -9\overset{\wedge }{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+\overset{\wedge }{\mathop{k}}\,is given by
(r→−i∧ +j∧ −2k∧ ).(−9i∧ −3j∧ +k∧ )=0 ⇒r→.(−9i∧ −3j∧ +k∧ )−(i∧ +j∧ −2k∧ ).(−9i∧ −3j∧ +k∧ )=0 ⇒r→.(−9i∧ −3j∧ +k∧ )=(i∧ +j∧ −2k∧ ).(−9i∧ −3j∧ +k∧ ) ⇒r→.(−9i∧ −3j∧ +k∧ )=−9−3−2 ⇒r→.(−9i∧ −3j∧ +k∧ )=−14...............(1) \begin{aligned} & \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{aligned}
Now, consider the given equation of the line, r→=3i∧ −j∧ −k∧ +λ(2i∧ −2j∧ +k∧ )...............(2)\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,
(3i∧ −j∧ −k∧ +λ(2i∧ −2j∧ +k∧ )).(−9i∧ −3j∧ +k∧ )=−14 ⇒((3+2λ)i∧ +(−1−2λ)j∧ +(−1+λ)k∧ ).(−9i∧ −3j∧ +k∧ )=−14 ⇒−27−18λ+3+6λ−1+λ=−14 ⇒−11−11λ=0 ⇒−11λ=11 ⇒λ=−1111 ⇒λ=−1 \begin{aligned} & \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{aligned}
Now, by substituting the value of λ\lambda in equation (2), we get,
r→=3i∧ −j∧ −k∧ +(−1)(2i∧ −2j∧ +k∧ ) ⇒r→=3i∧ −j∧ −k∧ −2i∧ +2j∧ −k∧  ⇒r→=i∧ +j∧ −2k∧  \begin{aligned} & \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{aligned}

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

Note: There is a possibility of making a mistake by considering AB→=A→−B→\overrightarrow{AB}=\overrightarrow{A}-\overrightarrow{B} instead of AB→=B→−A→\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 n→=AB→×BC→\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{BC} instead of n→=AB→×AC→\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.