Resnick Halliday Krane Unit Vectors question

[Neek 007]

The start of a new school year in Ap physics C, and our favorite engineering professors causing students distress! Maybe the champions of the textbook can help me out... = )

Homework Statement

1. Two vectors are given by
a = 4i hat - 3j hat + k hat
b = -i hat +j hat + 4k hat.

Find
a) a + b
b) a-b
C) vector c such that a -b +c = 0

2.Given two vectors
a = 4i hat - 3j hat
b = 6i hat + 8j hat

Find the magnitudes and directions (with the +x axis) of

a)a
b)b
c)a+b
d)b-a
e)a-b

Homework Equations

components equations
etc

The Attempt at a Solution

on question 1, I've gotten parts a and b, but running into difficulty proving a-b + c=0.
i took a-b from part b and replaced it into the equation.

C)
a - b = 3i hat + 2j hat

3i hat + 2j hat + c = 0

3i hat + 2j hat + (a_zk hat + b_z k hat)

3i hat + 2j hat + (1 + 4)k hat

3i hat + 2j hat + 5k hat = 0

i solved for vector c as 5k hat, but i am not sure if this satisfies the solution, as k hat runs into the z direction.
on question 2,

e) a - b

a = 4i hat -3j hat
b= 6i hat +8j hat

a-b=-2i hat - 11j hat

-2i hat = c_x = -2
-11j hat = c_y = -11

\sqrt{125} = 11.180

tan \Phi = (-11/-2) = 79.650 degrees

vector c = 11.180 at 79.650 degrees with +x-axis.

this is the answer i had gotten, but the book answer decides to choose 260 degrees. i have looked at my positive and negative symbols for my angle measure, but cannot meet the same number.

Any help is much appreciated :)

 

[Doc Al]

on question 1, I've gotten parts a and b, but running into difficulty proving a-b + c=0.
i took a-b from part b and replaced it into the equation.

C)
a - b = 3i hat + 2j hat

Redo that subtraction.

on question 2,

e) a - b

a = 4i hat -3j hat
b= 6i hat +8j hat

a-b=-2i hat - 11j hat

-2i hat = c_x = -2
-11j hat = c_y = -11

\sqrt{125} = 11.180

tan \Phi = (-11/-2) = 79.650 degrees

vector c = 11.180 at 79.650 degrees with +x-axis.

Hint: What quadrant should your answer be in?

 

[Clever-Name]

For Question 1 C consider it to be a normal algebraic equation. If it were asking you for a number, z, such that if x = 3, y = 5, x - y + z = 0 What would you do? You seem to be thinking that the vector c has to be the z component of this new a-b vector, but it isn't! It is its own unique vector \vec{c} = c_{x}\hat{i} + c_{y}\hat{j} + c_{z}\hat{k}

 

[cepheid]

Keep in mind that to add two vectors that are expressed in component form, you add them component-wise. In other words, the x-component of the resultant is the sum of the x-components of all the individual vectors, and likewise for y and z. Bearing that in mind, I think the best approach for part C is to use a vector c = (c_x, c_y, c_z) whose components are unknown (note: boldface denotes vectors). This way you have three unknowns, but you also have three equations (which come from the component-wise summation) so you can solve for them all.

 

[Neek 007]

Great! I got both of them. for question 1 it looks like i just got a bit lazy. question 2 i did indeed needed to refer to the quadrant. For my future reference, i should always check which quadrant i am going to end up in at the end.

oh yea another small question. What does "with + x axis" mean? does it mean with respect to the x axis?

Thanks a lot!

 

[Doc Al]

oh yea another small question. What does "with + x axis" mean? does it mean with respect to the x axis?

Yes. (As opposed to the negative x axis.)

 

[mikelepore]

For my future reference, i should always check which quadrant i am going to end up in at the end.

A handy rule to remember is: When taking an inverse tangent on a calculator, if the answer is in the 2nd or 3rd quadrant, add 180 degrees to the answer that the calculator gives you.