Question:

For the vectors shown in Fig. 3-35, determine

- $\vec{B} - 3\vec{A}$,
- $2\vec{A} - 3\vec{B} + 2\vec{C}$.

Figure 3-35Vector magnitudes are given in arbitrary units.

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

- $137\textrm{ units, } 16.9^\circ \textrm{ below neg. x-axis}$
- $149.7\textrm{ units, } 35.3^\circ \textrm{ below pos. x-axis}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko.*B*minus 3

*A*is the same as the vector

*B*plus negative 3 times

*A*and the reason I like to think of it that way is because you are still adding vectors so it's still the head to tail method and we put on the head of vector

*B*— head of the first vector— we put the tail of the second vector and the second vector in this case is gonna be vector

*A*but in the opposite direction, that's what the minus sign there means, and so here's negative 3

*A*it's 3 times the length in the opposite direction and it's still 28 degrees but this time it's 28 degrees below the

*x*-axis and our resultant will go from the very beginning which is at the tail of vector

*B*to the very end which is the head of vector

*A*or 3

*A*I guess. The

*x*-component of the resultant is gonna be the sum of the

*x*-components of these vectors we are working with. So we have negative 3 times

*A x*plus

*B x*and the

*x*-component of

*A*is gonna be 44— it's length times the

*cos*of 28— and then that's gonna be negative because it's to the left and then minus 26.5 which is the length of

*B*times

*cos*of 56 and that's a negative because it's to the left here and we get negative 131.37 is the

*x*-component of the resultant and that's this part—starting from here all the way to here. And the

*y*-component of the resultant is negative 3 times the

*y*-component of vector

*A*so it's negative 3 times 44 times

*sin*28 and then add to that the positive

*y*-component of vector

*B*— that's this portion right here— and the end result of that is negative 40 so there's a small

*y*-component downwards for the resultant. And the length of the resultant then will be the square root of its

*x*-component squared plus its

*y*-component squared and so that gives 137.32 units and then the angle will be the inverse tangent of the opposite of

*R*, its

*y*-component and divided by the

*x*-component of the resultant and that's 137.32 there and that gives 137 units, 16.9 degrees below the negative

*x*-axis for the resultant there. In part (b), we have 2

*A*so that's vector

*A*in the same direction but 2 times the length and then plus negative 3

*B*so that's vector

*B*but in the down, to the right now 3 times the length and it's directed 56 degrees below the positive

*x*-axis now and then plus 2

*C*so vector

*C*is straight down and it's twice the length so it's 62 units and 3 times

*B*is 79.5 and 2 times

*A*is 88. And then we get to work finding components and using Pythagoras to find the length of the resultant and use inverse tangent to get the angle, so I guess a little more slowly would be 2 times

*A x*is 88 and then

*cos*of 28 to get the

*x*-component of that plus 79.5 is the length of 3

*B*times the

*cos*of 56 gives the

*x*-component of

*B*, that's this part just in here that's

*x*-component of 3

*B*and then there's no

*x*-component for

*C*so that's just zero and we have 122.155 is the

*x*-component of the resultant and so that takes us this

*x*-component of the resultant would be like this so that's

*R x*. And then

*R y*is this full part from down there and

*R y*is found by going 2 times

*A y*plus 3

*B y*plus 2

*C y*;

*y*-component of

*A*is upwards and so it's 88 times

*sin*of 28 that's positive and this is the

*y*-component of

*A*right here 2

*A y*I suppose and then we have plus 3

*B y*now— this is in the downward direction so we put a minus sign in front of that— and so that's

*sin*of 56 and that gives us the opposite here and times 79.5 and then minus 62—that's twice the length of

*C*— that gives a negative 86.595. So the length of the resultant then is the square root of the

*x*-component— 122.155—squared plus the

*y*-component— negative 86.595—squared and that gives 149.7 units and then find the angle by going inverse tangent of the

*y*-component divided by the

*x*-component and then use the diagram to see that this is below the positive

*x*-axis. So that's 35.3 degrees below the positive

*x*-axis.

## Comments

why is he not multiplying the 3 to the A in part A as he does in B. In B he doubles and triples the lengths of the sides.

Hi olever8, thanks for the question. In part A, the video shows the vector A length multiplied by 3, as expected. Perhaps what you're noticing is that I overlooked saying it was multiplied by 3 at one point, but the multiplying does indeed happen.