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Q35E

Expert-verifiedFound in: Page 550

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.

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

**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 **\({\bf{v}}\)**.**

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.

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.

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.

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

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