# Thread: Vector Magnitude and Angle at 0x

1. ## Vector Magnitude and Angle at 0x

Sorry if this isn't pre-calc, the education system over here is different =\ .

Just checking if I have this right as our instructions and teachings were a bit vague.

Question: Given that a = i + j and b = 2i + j, find the magnitude of each of the following vectors and the angle each makes with Ox.

(d) : 3a - 2b

Magnitude : sqrt(26)
Angle: 101.3 degrees (3sf)

2. Originally Posted by Viral
Question: Given that a = i + j and b = 2i + j, find the magnitude of each of the following vectors and the angle each makes with Ox.
(d) : 3a - 2b
You need to rework the problem because $3a-2b=-i+j$.

3. Originally Posted by Plato
You need to rework the problem because $3a-2b=-i+j$.
What do you mean rework?

From what I can see, the lengh on the x axis would be 1, while the height on the y-axis would be 1 as well. However, as it is -1i, it would be on the negative x axis. So does this make a difference to the normal way of working it out?

I did:

3(i+j) - 2(2i-j)
= (3i+3j) - (4i - 2j)
= |-i + 5j|
= sqrt(i^2 + 5^2)
= sqrt(26)
angle = 180 - tan^-1(5)
angle = 101.3 degrees

The problem is, I'm not sure if I was supposed to do the 180- part or not?

4. Originally Posted by Viral
What do you mean rework?

From what I can see, the lengh on the x axis would be 1, while the height on the y-axis would be 1 as well. However, as it is -1i, it would be on the negative x axis. So does this make a difference to the normal way of working it out?

I did:

3(i+j) - 2(2i-j)
= (3i+3j) - (4i - 2j)
= |-i + 5j|
= sqrt(i^2 + 5^2)
= sqrt(26)
angle = 180 - tan^-1(5)
angle = 101.3 degrees

The problem is, I'm not sure if I was supposed to do the 180- part or not?
The problem says that $\color{blue}b=2i\large+j$

5. Hmm that was a typo in the first post.

$a = i + j$
$b = 2i - j$