Question 8 Exercise 3.5

Solutions of Question 8 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.

Find the volume of tetrahedron with the Vectors as coterminous edges a=ˆi+2ˆj+3ˆk,b=4ˆi+5ˆj+6ˆk,c=7ˆj+8ˆk

The volume of tetrahedron is \begin{align}V&=\dfrac{1}{6}[\vec{u} \cdot \vec{v} \times \vec{w}]\\ \Rightarrow V&=\dfrac{1}{6}\left|123456078\right|\\ V&=\dfrac{1}{6} \cdot 1(40-42)-4(16-21) \\ \Rightarrow V&=\dfrac{1}{6}(-2+20)=3 \text { units. }\end{align}

Find the volume of tetrahedron with A(2,3,1),B(1,2,0), C(0.2,5).D(0.1,2) as vertices.

Position vector of A,OA=2ˆi+3ˆj+ˆk

Position vector of B,OB=ˆi2ˆj

Position vector of C,OC=2ˆj5ˆk

Position vector of D,OD=ˆj2ˆk

We find the edges vectors \begin{align}\vec{a}&=\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\ & =(-\hat{i}-2 \hat{j})-(2 \hat{i}-3 \hat{j}+\hat{k}) \\ \Rightarrow \vec{a}&=-3 \hat{i}-5 \hat{j}-\hat{k} \\ \vec{b}&=\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A} \\ & =2 \hat{j}-5 \hat{k}-(2 \hat{i}+3 \hat{j}+\hat{k}) \\ \therefore \vec{b}&=2 \hat{i}-\hat{j} - 6 \hat{k} \\ \vec{c}&=\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A} \\ & =\hat{j}-2 \hat{k}-(2 \hat{i}+3 \hat{j}+\hat{k}) \\ \Rightarrow \vec{c}&=-2 \hat{i}-2 \hat{j}-3 \hat{k}\end{align} The volume of tetiahedron is: \begin{align}V&=\dfrac{1}{6} \left| 351216223 \right|\\ V& =\dfrac{1}{6}[ -3(3-12)+5(6-12)-1(4-2)]\\ & =\dfrac{1}{6}[ 27-30-2]\\ & \Rightarrow V=-\frac{5}{6} \text { units. } \end{align} Volume can not he negative, so V:56 units cube.