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Expert-verified Found in: Page 565 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (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 $${\bf{PQ}}R$$.

(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 $$\frac{1}{2}(\sqrt {86} )$$.

See the step by step solution

## 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 $$\frac{1}{2}(\sqrt {86} )$$. ### Want to see more solutions like these? 