# Sum of vectors

1. Jan 28, 2005

### Kp0684

What is the sum of the following four vectors in (a) unit- vector notation, and as (b) a magnitude and (c) an angle?........... A=(2.00m)i + (3.00m)j....B: 4.00m, at +65.0 degrees.......C= (-4.00m)i - (6.00m)j......D: 5.00m, at -235 degrees........i understand how to get the magnitude and the angle but how would i set this up......... would i start with A and C and find their sum.........i believe whats confusing me is B and D......otherwise i know how to set it up......need help.....

2. Jan 28, 2005

### dextercioby

HINT:Write all 4 vectors in component form.

Then simply add the 4 numbers for the "x" component and the 4 numbers for the "y" component.

Daniel.

3. Jan 28, 2005

### marlon

Here are the formula's you will need to apply.

Given two vectors with components A = (i,j,k) and B = (a,b,c)

magnitude $$\sqrt{a^2 + b^2 + c^2}$$
scalar product: A.B = ia + b j + ck
scalar product: A.B = magnitude of A * magnitude of B * cos(t) where t is the angle between the two vectors A and B

sum A+B = (i+a,j+b,k+c) (this is a new vector, the scalar product yields a number)
marlon

4. Jan 28, 2005

### Kp0684

okay, i get -2.00i - 3.00j - 1.17k when i sum it up.......iam still lost on this one....need help again.......

5. Feb 4, 2005

### Paloseco

i, j and k represent the 3 directions of the space, for example high, wide and length. The object will be placed this 3 distances in the space from the point you consider as reference

6. Feb 4, 2005

### HallsofIvy

Staff Emeritus
I'm very confused! Where did "k" come from? You original post said
"......... A=(2.00m)i + (3.00m)j....B: 4.00m, at +65.0 degrees.......C= (-4.00m)i - (6.00m)j......D: 5.00m, at -235 degrees........" with only two dimensions.

I presume that the angles are measured relative to the positive x axis.
B would be 4cos(65)i+ 4sin(65)j and D would be 5cos(-235)i+ 5sin(-235)j

Adding those four vectors does not give you any "k" component.

Of course, once you have found the sum, you find the length by the Pythagorean theorem and the angle is arctan((j component)/(x component)).

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