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Q35E

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Found in: Page 565

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# To find: The volume of the parallelepiped determined with adjacent edges PQ, PR and PS.

The volume of the parallelepiped with the adjacent edges PQ, PR and PS is $$16$$ cubic units.

See the step by step solution

## Step 1: Concept of parallelepiped with the help of Formula’s

Consider two three-dimensional vectors such as,

$$\begin{array}{l}a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \\{\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \end{array}$$

Write the expression for cross product between $$a$$ and $${\rm{b}}$$ vectors.

$$a \times b = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|(1)$$

Write the expression for dot product between $$a$$ and $${\rm{b}}$$ vectors.

$$a \cdot b = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}(2)$$

Write the expression to calculate the parallelepiped's volume.

$$V = |{\rm{a}} \cdot ({\rm{b}} \times {\rm{c}})|(3)$$

## Step 2: Volume of parallelepiped by calculating through Formula.

$$P = ( - 2,1,0),Q = (2,3,2){\rm{ and R}} = (1,4, - 1)andS = (3,6,1)$$

Consider the adjacent edges as $${\rm{a}} = \overrightarrow {PQ} ,\;{\rm{b}} = \overrightarrow {PR}$$ and $${\rm{c}} = \overrightarrow {PS}$$

Find $${\rm{a}} = \overrightarrow {PQ}$$.

$$\begin{array}{l}\overrightarrow {PQ} = Q(2,3,2) - P( - 2,1,0)\\\overrightarrow {PQ} = \langle (2 + 2),(3 - 1),(2 - 0)\rangle \\\overrightarrow {PQ} = \langle 4,2,2\rangle \end{array}$$

Find $${\rm{b}} = \overrightarrow {PR}$$.

$$\begin{array}{l}\overrightarrow {PR} = R(1,4, - 1) - P( - 2,1,0)\\\overrightarrow {PR} = \langle (1 + 2),(4 - 1),( - 1 - 0)\rangle \\\overrightarrow {PR} = \langle 3,3, - 1\rangle \end{array}$$

Find $${\rm{c}} = \overrightarrow {PS}$$.

$$\begin{array}{l}\overrightarrow {PS} = S(3,6,1) - P( - 2,1,0)\\\overrightarrow {PS} = \langle (3 + 2),(6 - 1),(1 - 0)\rangle \\\overrightarrow {PS} = \langle 5,5,1\rangle \end{array}$$

Modify equation ($$1$$).

$$b \times c = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|$$

We can substitute $$3$$ for $${b_1},3$$ for $${b_2}, - 1$$ for $${b_3},5$$ for $${c_1},1$$ for $${c_2}$$ and $$1$$ for $${c_3}$$,

$$\begin{array}{l}b \times c = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\3&3&{ - 1}\\5&5&1\end{array}} \right|\\b \times c = \left| {\begin{array}{*{20}{l}}3&{ - 1}\\5&1\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{l}}3&{ - 1}\\5&1\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{l}}3&3\\5&5\end{array}} \right|{\rm{k}}\\b \times c = (3 + 5){\rm{i}} - (3 + 5){\rm{j}} + (15 - 15){\rm{k}}\\b \times c = 8{\rm{i - }}8{\rm{j}} - 0{\rm{k}}\end{array}$$S

Modify equation $$\left( 2 \right)$$

$$a \cdot (b \times c) = {a_1}\left( {{b_1} \times {c_1}} \right) + {a_2}\left( {{b_2} \times {c_2}} \right) + {a_3}\left( {{b_3} \times {c_3}} \right)$$

We are substituting $$4$$ for $${a_1},2$$ for $${a_2},2$$ for $${a_3},8$$ for $$\left( {{b_1} \times {c_1}} \right), - 8$$ for $$\left( {{b_2} \times {c_2}} \right)$$ and $$0$$ for $$\left( {{b_3} \times {c_3}} \right)$$,

$$\begin{array}{l}a \cdot (b \times c) = 4(8) + 2( - 8) + 2(0)\\a \cdot (b \times c) = 32 - 16 + 0\\a \cdot (b \times c) = 16\end{array}$$

We are substituting $$16$$ for $$a \cdot (b \times c)$$ in equation $$(3)$$

$$\begin{array}{l}V = |16|\\V = 16\end{array}$$

Thus, the volume of the parallelepiped determined by the vectors $$a$$, $$b$$ and $$c$$ is $$16$$ cubic units.