2.3 Algebra of vectors (Page 4/8)

University physics volume 1 Page 4 / 8

Vector algebra

Find the magnitude of the vector

\vec{C}

that satisfies the equation

2 \vec{A} - 6 \vec{B} + 3 \vec{C} = 2 \hat{j}

, where

\vec{A} = \hat{i} - 2 \hat{k}

and

\vec{B} = − \hat{j} + \hat{k} / 2

Strategy

We first solve the given equation for the unknown vector

\vec{C}

. Then we substitute

\vec{A}

and

\vec{B}

; group the terms along each of the three directions

\hat{i}

\hat{j}

, and

\hat{k}

; and identify the scalar components

C_{x}

C_{y}

, and

C_{z}

. Finally, we substitute into [link] to find magnitude C .

Solution

\begin{matrix} 2 \vec{A} - 6 \vec{B} + 3 \vec{C} & = & 2 \hat{j} \\ 3 \vec{C} & = & 2 \hat{j} - 2 \vec{A} + 6 \vec{B} \\ \vec{C} & = & \frac{2}{3} \hat{j} - \frac{2}{3} \vec{A} + 2 \vec{B} \\ = & \frac{2}{3} \hat{j} - \frac{2}{3} (\hat{i} - 2 \hat{k}) + 2 (− \hat{j} + \frac{\hat{k}}{2}) = \frac{2}{3} \hat{j} - \frac{2}{3} \hat{i} + \frac{4}{3} \hat{k} - 2 \hat{j} + \hat{k} \\ = & - \frac{2}{3} \hat{i} + (\frac{2}{3} - 2) \hat{j} + (\frac{4}{3} + 1) \hat{k} \\ = & - \frac{2}{3} \hat{i} - \frac{4}{3} \hat{j} + \frac{7}{3} \hat{k} . \end{matrix}

The components are $C_{x} = − 2 / 3$ , $C_{y} = −4 / 3$ , and $C_{z} = 7 / 3$ , and substituting into [link] gives

C = \sqrt{C_{x}^{2} + C_{y}^{2} + C_{z}^{2}} = \sqrt{{(−2 / 3)}^{2} + {(− 4 / 3)}^{2} + {(7 / 3)}^{2}} = \sqrt{23 / 3} .

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Displacement of a skier

Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwest before taking a rest. Find his total displacement vector relative to the lodge when he is at the rest point. How far and in what direction must he ski from the rest point to return directly to the lodge?

Strategy

We assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector

\hat{i}

pointing east and the unit vector

\hat{j}

pointing north. There are three displacements:

{\vec{D}}_{1}

{\vec{D}}_{2}

, and

{\vec{D}}_{3}

. We identify their magnitudes as

D_{1} = 5.0 km

D_{2} = 3.0 km

, and

D_{3} = 4.0 km

. We identify their directions are the angles

θ_{1} = 90 °

θ_{2} = 180 °

, and

θ_{3} = 180 ° + 45 ° = 225 °

. We resolve each displacement vector to its scalar components and substitute the components into [link] to obtain the scalar components of the resultant displacement

\vec{D}

from the lodge to the rest point. On the way back from the rest point to the lodge, the displacement is

\vec{B} = − \vec{D}

. Finally, we find the magnitude and direction of

\vec{B}

Solution

Scalar components of the displacement vectors are

\begin{matrix} {\begin{matrix} D_{1 x} = D_{1} cos θ_{1} = (5.0 km) cos 90 ° = 0 \\ D_{1 y} = D_{1} sin θ_{1} = (5.0 km) sin 90 ° = 5.0 km \end{matrix} \\ {\begin{matrix} D_{2 x} = D_{2} cos θ_{2} = (3.0 km) cos 180 ° = −3.0 km \\ D_{2 y} = D_{2} sin θ_{2} = (3.0 km) sin 180 ° = 0 \end{matrix} \\ {\begin{matrix} D_{3 x} = D_{3} cos θ_{3} = (4.0 km) cos 225 ° = −2.8 km \\ D_{3 y} = D_{3} sin θ_{3} = (4.0 km) sin 225 ° = −2.8 km \end{matrix} \end{matrix} .

Scalar components of the net displacement vector are

{\begin{matrix} D_{x} = D_{1 x} + D_{2 x} + D_{3 x} = (0 - 3.0 - 2.8) km = −5.8 km \\ D_{y} = D_{1 y} + D_{2 y} + D_{3 y} = (5.0 + 0 - 2.8) km = + 2.2 km \end{matrix} .

Hence, the skier’s net displacement vector is $\vec{D} = D_{x} \hat{i} + D_{y} \hat{j} = (−5.8 \hat{i} + 2.2 \hat{j}) km$ . On the way back to the lodge, his displacement is $\vec{B} = − \vec{D} = − (−5.8 \hat{i} + 2.2 \hat{j}) km = (5.8 \hat{i} - 2.2 \hat{j}) km$ . Its magnitude is $B = \sqrt{B_{x}^{2} + B_{y}^{2}} = \sqrt{{(5.8)}^{2} + {(−2.2)}^{2}} km = 6.2 km$ and its direction angle is $θ = {tan}^{−1} (−2.2 / 5.8) = −20.8 °$ . Therefore, to return to the lodge, he must go 6.2 km in a direction about $21 °$ south of east.

Significance

Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using a graphical method; however, we can check if our solution makes sense by sketching it, which is a useful final step in solving any vector problem.

Displacement of a jogger

A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 m before he stops at a drinking fountain ( [link] ). His displacement vector from point A at the bottom of the steps to point B at the fountain is

{\vec{D}}_{A B} = (−90.0 \hat{i} + 30.0 \hat{j}) m

. What is the height and width of each step in the flight? What is the actual distance the jogger covers? If he makes a loop and returns to point A , what is his net displacement vector?

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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