====== 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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