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Essential Calculus: Early Transcendentals
Found in: Page 565
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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.

Most popular questions for Math Textbooks

(a) To determine

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(b) To determine

To find: The area of triangle \({\bf{PQ}}R\).

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(b) To sketch the summation vector \({\bf{c}} = s{\bf{a}} + t{\bf{b}}\).

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