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Q35E

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Essential Calculus: Early Transcendentals
Found in: Page 550
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

(a) To sketch the vectors \({\rm{a}} = \langle 3,2\rangle ,b = \langle 2, - 1\rangle \), and \({\rm{c}} = \langle 7,1\rangle \).

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

(c) To estimate the values of \(s\)and \(t\) using sketch.

(d) To find the exact values of \(s\)and \(t\).

a) The vectors \({\bf{a}},{\bf{b}}\) and \({\bf{c}}\) is sketched.

b) The vector \({\bf{c}} = s{\bf{a}} + t{\bf{b}}\) is sketched.

c) The values of \(s\)and \(t\) are \(\underline {1.3} \) and \(\underline {1.6} \) respectively.

d) The exact values of \(s\)and \(t\) are \(\frac{9}{7}\) and \(\frac{{11}}{7}\) respectively.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given

Two-dimensional vectors \({\bf{a}}\) is \(\langle 3,2\rangle ,\;{\bf{b}}\) is \(\langle 2, - 1\rangle \), and \({\bf{c}}\) is \(\langle 7,1\rangle \).

Step 2: Concept of Scalar multiplication

Scalar multiplication:

Consider a scalar \(c\) and \(a\) vector \({\bf{v}}\).

The scalar multiple \(c{\bf{v}}\) which is \(a\) vector with a length more than \(|c|\)times of vector \({\bf{v}}\) in same direction of \({\bf{v}}\).

Step 3: Draw the Vectors

a)

The three two-dimensional vectors lie on \(xy\)-plane.

Plot the point \((3,2)\) in \(xy\)-plane and connect the origin to point through line with origin to point direction this forms a vector \({\bf{a}} = \langle 3,2\rangle \).

Plot the point \((2, - 1)\) in \(xy\)-plane and connect the origin to point through line with origin to point direction this forms a vector \({\bf{b}} = \langle 2, - 1\rangle \).

Plot the point \((7,1)\) in \(xy\)-plane and connect the origin to point through line with origin to point direction this forms a vector \({\bf{c}} = \langle 7,1\rangle \).

From explanation, draw the vectors as shown in Figure 1.

Thus, the vectors \({\bf{a}},\;{\bf{b}},\;{\bf{c}}\) are sketched.

Step 4: Draw the Summation Vectors

b)

The three two-dimensional vectors lie on \(xy\)-plane.

Plot the point \((3,2)\) in \(xy\)-plane and connect the origin to point through line with origin to point direction this forms a vector \({\bf{a}} = \langle 3,2\rangle \).

Plot the point \((2, - 1)\) in \(xy\)-plane and connect the origin to point through line with origin to point direction this forms a vector \({\bf{b}} = \langle 2, - 1\rangle \).

Plot the point \((7,1)\) in \(xy\)-plane and connect the origin to point through line with origin to point direction this forms a vector \({\bf{c}} = \langle 7,1\rangle \).

Extend the vectors \({\bf{a}}\) and \({\bf{b}}\) in their respective directions.

Connect a line from the terminal point of vector \({\bf{c}}\) to extended vectors of \({\bf{a}}\) and \({\bf{b}}\).

This extended part of vector \({\bf{a}}\) is scalar \(s\) and extended part of \({\bf{b}}\) is scalar \(t\).

This is the sketch of \({\bf{c}} = s{\bf{a}} + t{\bf{b}}\).

From explanation, draw the sketch of \({\bf{c}} = s{\bf{a}} + t{\bf{b}}\) as shown in Figure 2.

Thus, the vector \({\bf{c}} = s{\bf{a}} + t{\bf{b}}\) is sketched.

Step 5: Calculate the values by the Diagram

c)

From Figure 2, the length of vectors \(sa\) and \(tb\) are\(4\)and \(3.2\) respectively.

From Figure 1, the length of vectors \({\bf{a}}\) and \({\bf{b}}\) are \(3\)and \(2\) respectively.

In accordance with the scalar multiplication, the scalar value is determined by the ratio of length of scalar multiple and length of vector.

Consider the expression to find scalar \(s\).

\(s = \frac{{|s{\bf{a}}|}}{{|{\bf{a}}|}}\)

Substitute \(4\)for \(sa\) and \(3\)for \({\bf{a}}\).

\(\begin{aligned}{c}s &= \frac{4}{3}\\ &= 1.3\end{aligned}\)

Consider the expression to find scalar \(t\).

\(t = \frac{{|{\bf{b}}|}}{{|{\bf{b}}|}}\)

Substitute \(3.2\) for \(sa\)and \(2\)for \({\bf{a}}\).

\(\begin{aligned}{c}t &= \frac{{3.2}}{2}\\ &= 1.6\end{aligned}\)

Thus, the values of \(s\)and \(t\)are \(\underline {1.3} \) and \(\underline {1.6} \) respectively.

Step 6: Calculate the accurate values

d)

Write the expression for vector \({\bf{c}}\).

\({\bf{c}} = s{\bf{a}} + t{\bf{b}}\) … (1)

The \(x\) and \(y\)-coordinates of vector \(c\) are estimated by the use of equation (1).

Substitution of \(x\)-coordinated of vectors \({\bf{a}}\) and \({\bf{b}}\) along with scalars results \(x\)-coordinate of vector \({\bf{c}}\) and substitution of \(y\)-coordinates of vectors \({\bf{a}}\) and \({\bf{b}}\)along with scalars provides the value of \(y\)-coordinate of vector \({\bf{c}}\).

Substitute \(3 \to {\bf{a}},\;2 \to {\bf{b}}\), and \(7 \to {\bf{c}}\)in equation (1).

\(\begin{aligned}{c}7 &= s(3) + t(2)\\3s + 2t &= 7\end{aligned}\) … (2)

Substitute \(2 \to {\bf{a}}, - 1 \to {\bf{b}}\), and \(1 \to {\bf{c}}\) in equation (1).

\(\begin{aligned}{c}1 &= s(2) + t( - 1)\\2s - t &= 1\\t &= 2s - 1\end{aligned}\) … (3)

Substitute \(2s - 1 \to t\) in equation (2).

\(\begin{aligned}{c}3s + 2(2s - 1) &= 7\\3s + 4s - 2 &= 7\\7s &= 7 + 2\\s &= \frac{9}{7}\end{aligned}\)

Substitute \(\frac{9}{7} \to s\) in equation (3).

\(\begin{aligned}{c}t &= 2\left( {\frac{9}{7}} \right) - 1\\ &= \frac{{18}}{7} - 1\\ &= \frac{{18 - 7}}{7}\\ &= \frac{{11}}{7}\end{aligned}\)

Thus, the exact values of \(s;\;t\) are \(\frac{9}{7};\;\frac{{11}}{7}\)respectively

Most popular questions for Math Textbooks

To find a dot product between \({\rm{a}}\) and \({\rm{b}}\).

To d\({\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)escribe all set of points for condition \(\left| {{\bf{r}} - {{\bf{r}}_0}} \right| = 1\).

To determine whether the given expression is meaningful or meaningless.

(a) \(({\rm{a}} \cdot {\rm{b}}) \cdot {\rm{c}}\)

(b) \((a \cdot b)c\)

(c) \(|{\rm{a}}|({\rm{b}} \cdot {\rm{c}})\)

(d) \(a \cdot (b + c)\)

(e) \(a \cdot b + c\)

(f) \(|a| \cdot (b + c)\)

To show

(a) \({\rm{i}} \cdot {\rm{j}} = 0,{\rm{j}} \cdot {\rm{k}} = 0\) and \({\rm{k}} \cdot {\rm{i}} = 0\).

(b) \({\rm{i}} \cdot {\rm{i}} = 1,{\rm{j}} \cdot {\rm{j}} = 1\) and \({\rm{k}} \cdot {\rm{k}} = 1\)

To find a dot product between \({\rm{a}}\) and \({\rm{b}}\).
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