# Thread: Vectors problems need help

1. ## Vectors problems need help

my teacher gave me this problems but i just can't figure out this one, please i really need to know this and if you can post Steps on how to do it will be awesome. Also i would rather see the procedure than a lot of explanation.

Given u=<3,4> w=<-2,2> and v=<-5,6> find the following:

u + v_____v - u_____4u - 3v_____||4u - 3v||

Also:

a unit vector in the direction of v

find the direction angle for -u

v*w (is v times w)

find the angle between v and w

show that the standard unit vectors i and j are orthogonal.

I know this long but i hope someone knows this answers, this is not homework to get points, I'm studying for a college test.

2. Originally Posted by Cyberman86
Given u=<3,4> w=<-2,2> and v=<-5,6> find the following:

u + v_____v - u_____4u - 3v_____||4u - 3v||
u + v = <3,4> + <-5,6> = <3-5,4+6>

Surely you can do the next one.

3. Originally Posted by Cyberman86
my teacher gave me this problems but i just can't figure out this one, please i really need to know this and if you can post Steps on how to do it will be awesome. Also i would rather see the procedure than a lot of explanation.

Given u=<3,4> w=<-2,2> and v=<-5,6> find the following:

u + v_____
Just add the x and y components of each vector together.

$\displaystyle \vec{u}+\vec{v}=\left<3+(-5),4+6\right>=\color{red}\boxed{\left<-2,10\right>}$

v - u_____
Subtract the corresponding x and y components:

$\displaystyle \vec{u}-\vec{v}=\left<3-(-5),4-6\right>=\color{red}\boxed{\left<8,-2\right>}$

4u - 3v_____
Multiply each vector by the corresponding scalar coefficient, and then subtract them:

$\displaystyle 4\vec{u}-3\vec{v}=\left<4(3)-3(-5),4(4)-3(6)\right>=\color{red}\boxed{\left<27,-2\right>}$

||4u - 3v||
Since we have $\displaystyle 4\vec{u}-3\vec{v}$, apply the equation to find the norm of the vector:

$\displaystyle \parallel 4\vec{u}-3\vec{v}\parallel=\sqrt{27^2+(-2)^2}=\color{red}\boxed{\sqrt{733}}$

Also:

a unit vector in the direction of v
first find $\displaystyle \parallel\vec{v}\parallel$

$\displaystyle \parallel\vec{v}\parallel=\sqrt{(-5)^2+6^2}=\sqrt{61}$

Thus, the unit vector $\displaystyle \hat u_v$ is $\displaystyle \color{red}\boxed{\hat u_v=-\frac{5}{\sqrt{61}}\hat i +\frac{6}{\sqrt{61}}\hat j}$

find the direction angle for -u
I'll get back to this one. Is the angle supposed to be measured with respect to the x axis? I want to say that the angle would be $\displaystyle \vartheta=\tan^{-1}\bigg(\frac{u_y}{u_x}\bigg)$...

v*w (is v times w)
I'm assuming you mean find the dot product?

When you take the dot product, you're left with a scalar value. You need to multiply the corresponding x and y components together and then add what you have left over:

$\displaystyle \vec{v}\cdot\vec{w}=(-5)(-2)+(6)(2)=\color{red}\boxed{22}$

find the angle between v and w
The equation for finding the angle between two vectors is this :

$\displaystyle \vartheta=\cos^{-1}\bigg(\frac{\vec{v}\cdot\vec{w}}{\parallel\vec{v }\parallel\parallel\vec{w}\parallel}\bigg)$

Thus, $\displaystyle \vartheta=\cos^{-1}\bigg(\frac{22}{\sqrt{61}\times\sqrt{8}}\bigg)\a pprox\color{red}\boxed{5.19^{o}}$

show that the standard unit vectors i and j are orthogonal.
To show that two vectors are orthogonal, you need show that the dot products of the two vectors is zero.

$\displaystyle \hat i \cdot \hat j=\left<1,0\right>\cdot\left<0,1\right>=(1)(0)+(0) (1)=\color{red}\boxed{0}$

I hope this helps!

--Chris

4. I guess we'll never know if Cyberman86 can do them. Oh, well.

5. after seeing Chris showing the process of the Problem i now can do it and its easier than i thought.