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Q36E

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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

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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.

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