Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

(a) To determine
To find: A nonzero vector orthogonal to the plane through the points \({\bf{P}}\), \({\bf{Q}}\) and \(R\).
(b) To determine
To find: The area of triangle \({\bf{PQ}}R\).

(a) The nonzero vector that runs through the points \[P\], \[Q\] and \[R\] is \(\langle - 1, - 7,6\rangle \) orthogonal to the plane.

(b) The area of triangle \(P{\rm{ }}Q{\rm{ }}R\) is \(\frac{1}{2}(\sqrt {86} )\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept of Nonzero vector, Orthogonal vectors, cross product and Area of Triangle with the help of Formula.

A vector with at least one non-zero entry, at least in Rn or Cn, is called a non-zero vector. In general, a non-zero vector is one that is not the vector space's identity element for addition.

Two vectors are said to be orthogonal if they are perpendicular to one another. The dot product of the two vectors, in other words, is zero.

Formula

(i) Write the expression for cross product between \[a\] and \[b\] vectors.

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

(ii) Consider the expression for area of the triangle in vector coordinate.

Area of triangle \( = \frac{1}{2}|a \times b|\left( 2 \right)\)

Step 2: calculating non zero orthogonal vector points with cross product formula.

(a)

\[P(0,0, - 3),Q(4,2,0)andR(3,3,1)\]

The vectors \[\overrightarrow {PQ} \times \overrightarrow {PR} \] is perpendicular to both \[\overrightarrow {PQ} \] and \[\overrightarrow {PR} \], therefore perpendicular to the plane through \[P(0,0, - 3),Q(4,2,0)\] and \[R(3,3,1)\].

Find \[\overrightarrow {PQ} \].

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

Find \(\overrightarrow {PR} \).

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

Re-Modify equation (1).

\[\begin{array}{l}\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{l}}2&3\\3&4\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}4&3\\3&4\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}4&2\\3&3\end{array}} \right|{\rm{k}}\\\overrightarrow {PQ} \times \overrightarrow {PR} = (8 - 9){\rm{i}} - (16 - ){\rm{j}} + (12 - 6){\rm{k}}\\\overrightarrow {PQ} \times \overrightarrow {PR} = - 1{\rm{i - 7j}} + 6{\rm{k}}\\\overrightarrow {PQ} \times \overrightarrow {PR} = \langle - 1, - 7,6\rangle \end{array}\]

Thus, the nonzero vector orthogonal to the plane through the points \(P\), \(Q\) and \(R\) is \(\langle - 1, - 7,6\rangle \).

Step 3: calculating area of Triangle with the help of Formula.

(b)

Here, \({\rm{a}}\) and \({\bf{b}}\) are the vectors.

Re-modify equation\(\left( 2 \right)\).

Area of triangle \( = \frac{1}{2}|\overrightarrow {PQ} \times \overrightarrow {PR} |\)

We can Substituting, \(\langle - 1, - 7,6\rangle \) for \(\overrightarrow {PQ} \times \overrightarrow {PR} \),

Area of triangle \( = \frac{1}{2}|\langle 0,18, - 9\rangle |\)

\[\begin{array}{l} = \frac{1}{2}\left( {\sqrt {{{( - 1)}^2} + {{( - 7)}^2} + {{(6)}^2}} } \right)\\ = \frac{1}{2}(\sqrt {1 + 49 + 36} )\\ = \frac{1}{2}(\sqrt {86} )\end{array}\]

Thus, the area of triangle \(P{\rm{ }}Q{\rm{ }}R\) is \(\frac{1}{2}(\sqrt {86} )\).

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Essential Calculus: Early Transcendentals

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Short Answer

(a) To determine
To find: A nonzero vector orthogonal to the plane through the points \({\bf{P}}\), \({\bf{Q}}\) and \(R\).
(b) To determine
To find: The area of triangle \({\bf{PQ}}R\).

Step by Step Solution

Step 1: Concept of Nonzero vector, Orthogonal vectors, cross product and Area of Triangle with the help of Formula.

Step 2: calculating non zero orthogonal vector points with cross product formula.

Step 3: calculating area of Triangle with the help of Formula.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

(a) To determineTo find: A nonzero vector orthogonal to the plane through the points \({\bf{P}}\), \({\bf{Q}}\) and \(R\).(b) To determineTo find: The area of triangle \({\bf{PQ}}R\).

Step by Step Solution

Step 1: Concept of Nonzero vector, Orthogonal vectors, cross product and Area of Triangle with the help of Formula.

Step 2: calculating non zero orthogonal vector points with cross product formula.

Step 3: calculating area of Triangle with the help of Formula.

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(a) To determine
To find: A nonzero vector orthogonal to the plane through the points \({\bf{P}}\), \({\bf{Q}}\) and \(R\).
(b) To determine
To find: The area of triangle \({\bf{PQ}}R\).