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Found in: Page 550

### Essential Calculus: Early Transcendentals

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

# To determine A geometric argument to show the vector $${\bf{c}} = s{\bf{a}} + t{\bf{b}}$$.

The argument of vector $${\bf{c}}$$ in terms of nonzero and nonparallel vectors $${\bf{a}}$$ and $${\bf{b}}$$ is $$\underline {\bf{c}} = s{\bf{a}} + t{\bf{b}}$$ and arguments of $$s$$ and $$t$$ are $$\frac{{{b_2}{c_1} - {b_1}{c_2}}}{{\underline {{b_2}{a_1} - {b_1}{a_2}} }}$$ and $$\frac{{{c_2}{a_1} - {c_1}{a_2}}}{{\underline {{b_2}{a_1} - {b_1}{a_2}} }}$$ respectively.

See the step by step solution

## Step 1: Given data

Non zero and non-parallel two-dimensional vectors $${\bf{a}}$$ and $${\bf{b}}$$ and any vector $${\bf{c}}$$.

## 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

Consider two-dimensional vectors $${\bf{a}} = \left\langle {{a_1},{a_2}} \right\rangle ,{\bf{b}} = \left\langle {{b_1},{b_2}} \right\rangle$$, and $${\bf{c}} = \left\langle {{c_1},{c_2}} \right\rangle$$.

The vectors are located in $$xy$$-plane as they are two-dimensional.

Sketch the vectors $${\bf{a}},{\bf{b}}$$, and $${\bf{c}}$$ in $$xy$$-plane from origin. Name the origin as 0.

Extend the vector $${\bf{a}}$$ by line $$A$$ and vector $${\bf{b}}$$ by $$B$$.

Draw the parallel lines $${A^\prime }$$ and $${B^\prime }$$ to vectors $${\bf{a}}$$ and $${\bf{b}}$$ from terminal point of vector $${\bf{c}}$$.

Name the intersection point of lines $$A$$and $${A^\prime }$$ as $$P$$ and intersection point of $$B$$ and $${B^\prime }$$ as $$Q$$.

From explanation, sketch of vectors is shown in Figure 1.

From Figure 1, write the expression for vector $${\bf{c}}$$.

$${\bf{c}} = \overrightarrow {OP} + \overrightarrow {OQ}$$

From Figure 1, write the expression for scalar $$s$$ by the use of definition.

$$s = \frac{{|\overrightarrow {OP} |}}{{|{\bf{a}}|}}$$

Here, $$|\overrightarrow {OP} |$$ is positive length of $$\overrightarrow {OP}$$.

$$|{\bf{a}}|$$ Is positive length of vector $${\bf{a}}$$.

From Figure 1, write the expression for scalar $$t$$ by the use of definition.

$$t = \frac{{|\overrightarrow {OQ} |}}{{|{\bf{b}}|}}$$

Here, $$|\overrightarrow {OQ} |$$ is positive length of $$\overrightarrow {OQ}$$ and $$|{\bf{b}}|$$ is positive length of vector $${\bf{b}}$$.

## Step 4: Calculate the Constants and substitute their values

From Figure 1, write the expression for vector $${\bf{c}}$$ in terms of vectors $${\bf{a}}$$ and $${\bf{b}}$$.

$${\bf{c}} = s{\bf{a}} + {\bf{b}}$$ … (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 scalar results $$x$$-coordinate of vector $${\bf{c}}$$ and substitution of $$y$$-coordinates of vectors $${\bf{a}}$$ and $${\bf{b}}$$ along with scalar provides the value of $$y$$-coordinate of vector $${\bf{c}}$$.

Substitute $${a_1} \to {\bf{a}},\;{b_1} \to {\bf{b}}$$, and $${c_1} \to {\bf{c}}$$ in equation (1).

$${c_1} = s{a_1} + t{b_1}$$ … (2)

Multiply $${a_2}$$ on both sides of equation (2).

\begin{aligned}{l}{c_1}{a_2} &= {a_2}\left( {s{a_1} + t{b_1}} \right)\\{c_1}{a_2} &= s{a_1}{a_2} + t{b_1}{a_2}\end{aligned} … (3)

Substitute $${a_2} \to {\bf{a}},\;{b_2} \to {\bf{b}}$$, and $${c_2} \to {\bf{c}}$$ in equation (1).

$${c_2} = s{a_2} + t{b_2}$$ … (4)

Multiply $${a_1}$$ on both sides of equation (4).

\begin{aligned}{l}{a_1}{c_2} &= {a_1}\left( {s{a_2} + t{b_2}} \right)\\{c_2}{a_1} &= s{a_1}{a_2} + t{b_2}{a_1}\end{aligned} … (5)

Subtract equation (3) from (5).

\begin{aligned}{c}{c_2}{a_1} - {c_1}{a_2} &= s{a_1}{a_2} + t{b_2}{a_1} - \left( {s{a_1}{a_2} + t{b_1}{a_2}} \right)\\{c_2}{a_1} - {c_1}{a_2} &= s{a_1}{a_2} - s{a_1}{a_2} + t{b_2}{a_1} - t{b_1}{b_1}\\{c_2}{a_1} - {c_1}{a_2} &= t\left( {{b_2}{a_1} - {b_1}{a_2}} \right)\\t &= \frac{{{c_2}{a_1} - {c_1}{a_2}}}{{{b_7}{a_1} - {b_1}{a_7}}}\end{aligned}

Multiply $${b_2}$$ on both sides of equation (2).

\begin{aligned}{l}{b_2}{c_1} &= {b_2}\left( {s{a_1} + t{b_1}} \right)\\{b_2}{c_1} &= s{b_2}{a_1} + t{b_1}{b_2}\end{aligned} … (6)

Multiply $${b_1}$$ on both sides of equation (4).

\begin{aligned}{l}{b_1}{c_2} &= {b_1}\left( {s{a_2} + t{b_2}} \right)\\{b_1}{c_2} &= s{b_1}{a_2} + t{b_1}{b_2}\end{aligned} … (7)

Subtract equation (7) from (6).

\begin{aligned}{c}{b_2}{c_1} - {b_1}{c_2} &= s{b_2}{a_1} + t{b_1}{b_2} - \left( {s{b_1}{a_2} + t{b_1}{b_2}} \right)\\{b_2}{c_1} - {b_1}{c_2} &= s{b_2}{a_1} + t{b_1}{b_2} - s{b_1}{a_2} - t{b_1}{b_2}\\{b_2}{c_1} - {b_1}{c_2} &= s\left( {{b_2}{a_1} - {b_1}{a_2}} \right)\\s &= \frac{{{b_2}{c_1} - {b_1}{c_2}}}{{{b_7}{a_1} - {b_1}{a_7}}}\end{aligned}

Thus, the argument of vector $${\bf{c}}$$ in terms of nonzero and nonparallel vectors $${\bf{a}}$$ and $${\bf{b}}$$ is $$\underline {\bf{c}} = s{\bf{a}} + t{\bf{b}}$$ and arguments of $$s$$ and $$t$$are $$\frac{{{b_2}{c_1} - {b_1}{c_2}}}{{\underline {{b_2}{a_1} - {b_1}{a_2}} }}$$ and $$\frac{{{c_2}{a_1} - {c_1}{a_2}}}{{\underline {{b_2}{a_1} - {b_1}{a_2}} }}$$ respectively.