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Expert-verified Found in: Page 550 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (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 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 ### Want to see more solutions like these? 