Question

Vector A points in the negative y direction and has a magnitude of 5 units. Vector B has twice the magnitude and points in the positive x direction tion. Find the direction and magnitude of (a) A + B (b) A - B and (c) B - A

Answer 1

Draw a diagram of the vectors A and B:

Using the unit vector notation:

`\begin{gathered} A=-5\hat{j} \\ B=10\hat{i} \end{gathered}`

Remember that when a vector is written in terms of its components:

`V=V_x\hat{i}+V_y\hat{j}`

Then, its magnitude is given by:

`|V|=\sqrt[]{V^2_x+V^2_y}`

And its direction is given by the following rule depending on the sign of the horizontal component:

`\begin{gathered} \theta=\tan ^(-1)((V_y)/(V_x))\text{ if }V_x>0 \\ \theta=180+\tan ^(-1)((V_y)/(V_x))\text{ if }V_x<0 \end{gathered}`

a) A+B

`\begin{gathered} A+B=(-5\hat{j)}+(10\hat{i}) \\ =10\hat{i}-5\hat{j} \end{gathered}`

Since the horizontal component is positive, then:

`\begin{gathered} |A+B|=\sqrt[]{(10)^2+(-5)^2} \\ =\sqrt[]{125} \\ =5\cdot\sqrt[]{5} \\ \approx11.18 \end{gathered}`

`\begin{gathered} \theta_(A+B)=\tan ^(-1)((-5)/(10)) \\ =-26.565\ldotsº \\ =333.43º \end{gathered}`

b) A-B

`\begin{gathered} A-B=(-5\hat{j})-(10\hat{i}) \\ =-10\hat{i}-5\hat{j} \end{gathered}`

Since the x component is negative, then:

`\begin{gathered} |A-B|=\sqrt[]{(-10)^2+(-5)^2} \\ =\sqrt[]{125} \\ =5\cdot\sqrt[]{5} \\ \approx11.18\ldots \end{gathered}`
`\begin{gathered} \theta_(A-B)=180º+\tan ^(-1)((-5)/(-10)) \\ =180º+26.565\ldotsº \\ =206.565\ldotsº \end{gathered}`

c) B-A

Notice that B-A is equal to -(A-B). Then, the magnitude is the same and the direction is the opposite (substract 180º from the direction of A-B to find the direction of B-A):

`\begin{gathered} |B-A|=5\cdot\sqrt[]{5}\approx11.18 \\ \theta_(B-A)=26.565\ldotsº \end{gathered}`