Question 6 Exercise 3.5

Solutions of Question 6 of Exercise 3.5 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Do the points $(4. 2.1)$, $(5,1,6)$, $(2.2,-5)$ and $(3.5 .0)$ lie in a plane?

Let we denote the given points with $A(4,-2,1), B(5,1,6)$. $C(2,2,-5)$ and $D(3,5.0)$

then Position vector of $A, \overrightarrow{O A}=4 \hat{i}-2 \hat{j}+\hat{k}$

Position vector of $B, \overrightarrow{O B}=5 \hat{i}+\hat{j}+6 \hat{k}$

Position vector of $C, \overrightarrow{O C}=2 \hat{i}+2 \hat{i}-5 \hat{k}$

Position vector of $D, \overrightarrow{O D}=3 \hat{i}+5 \hat{j}$.

Then \begin{align}\vec{a}&=\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\\ \Rightarrow \vec{a}&=(5 \hat{i}+\hat{j}+6 \hat{k})-(4 \hat{i}-2 \hat{j}+\hat{k}) \\ \Rightarrow \vec{a}&=\hat{i}+3 \hat{j}+5 \hat{k} \ldots \ldots \ldots \ldots \ldots(1)\\ \vec{b}&=\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A} \\ \Rightarrow \vec{b}&=2 \hat{i}+2 \hat{j} \quad 5 \hat{k}-(4 \hat{i}-2 \hat{j}-\hat{k}) \\ \Rightarrow \vec{b}&=-2 \hat{i}+4 \hat{j}-6 \hat{k} ....(2)\\ \vec{c}&=\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A} \\ \Rightarrow \vec{c}&=3 \hat{i}+5 \hat{j}-(4 \hat{i}-2 \hat{j}+\hat{k}) \\ \Rightarrow \vec{c}&=-\hat{i}+5 \hat{j}-\hat{k} ....(3) \end{align}
Since $\vec{a} \cdot \vec{b}$ and $\vec{c}$ are the vectors formed by the combination of given points such that in which all the points are used.

So we check

\begin{align}\vec{a} \cdot \vec{b} \times \vec{c}&=\left|\begin{array}{ccc} 1 & 3 & 5 \\ -2 & 4 & -6 \\ -1 & 7 & -1 \end{array}\right|\\ \vec{a} \cdot \vec{b} \times \vec{c}&=1(-4+42)- 3(2 -6) +5(-14+4) \\ \Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}&=38+12-50=0 .\end{align} Hence $\vec{a} \cdot \vec{b} \times \vec{c}$ is zero. Thus the given points lie in same plane.