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.
Question 8(i)
Find the volume of tetrahedron with the Vectors as coterminous edges \begin{align}\vec{a}&=\hat{i}+2 \hat{j}+3 \hat{k},\\ \vec{b}&=4 \hat{i}+5 \hat{j}+6 \hat{k}, \\ \vec{c}&=7 \hat{j}+8 \hat{k}\end{align}
Solution
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|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 0 & 7 & 8\end{array}\right|\\ V&=\dfrac{1}{6} \cdot 1(40-42)-4(16-21) \\ \Rightarrow V&=\dfrac{1}{6}(-2+20)=3 \text { units. }\end{align}
Question 8(ii)
Find the volume of tetrahedron with $A(2,3,1), B(-1,-2,0)$, $C(0.2,-5) . D(0.1,-2)$ as vertices.
Solution
Position vector of $A,\overrightarrow{O A}=2 \hat{i}+3 \hat{j}+\hat{k}$
Position vector of $B, \overrightarrow{O B}=-\hat{i}-2 \hat{j}$
Position vector of $C, \overrightarrow{O C}=2 \hat{j}-5 \hat{k}$
Position vector of $D, \overrightarrow{O D}=\hat{j}-2 \hat{k}$
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| \begin{array}{rrr} -3 & -5 & -1 \\ -2 & -1 & -6 \\ -2 & -2 & -3 \end{array} \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 $V: \dfrac{5}{6}$ units cube.
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