# Thread: Gr 12. Vector Question Involving Dot Product of Two Vectors(Part 3)

1. ## Gr 12. Vector Question Involving Dot Product of Two Vectors(Part 3)

Okay, the last one for today hopefully phew...!
I can guaranteed this one is DIFFICULT!!!

Three vectors x, y, and z satisfy x + y + z = 0. Calculate the value of $\displaystyle \vec x \cdot \vec y\ + \vec y \cdot \vec z\ + \vec z \cdot \vec x$, is |x| = 2, |y| = 3, and |z| = 4.

Lol learning how to use LateX here to see if it helps Will be learning more hopefully haha
Thanks

2. Originally Posted by narutoblaze
Okay, the last one for today hopefully phew...!
I can guaranteed this one is DIFFICULT!!!
Okay, let's say the vectors have the following components:

$\displaystyle \textbf x = x_1\textbf i+x_2\textbf j+x_3\textbf k$

$\displaystyle \textbf y = y_1\textbf i+y_2\textbf j+y_3\textbf k$

$\displaystyle \textbf z = z_1\textbf i+z_2\textbf j+z_3\textbf k$

Three vectors x, y, and z satisfy x + y + z = 0.
This tells us that

$\displaystyle \left\{\begin{array}{c} x_1+y_1+z_1=0\\ x_2+y_2+z_2=0\\ x_3+y_3+z_3=0 \end{array}\right.$

|x| = 2, |y| = 3, and |z| = 4
Using the dot product property $\displaystyle \textbf u\cdot\textbf u=\lVert\textbf u\rVert^2,$ we have

$\displaystyle \left\{\begin{array}{rcl} x_1^2+x_2^2+x_3^2 & = & 4\\ y_1^2+y_2^2+y_3^2 & = & 9\\ z_1^2+z_2^2+z_3^2 & = & 16 \end{array}\right.$

Calculate the value of $\displaystyle \vec x \cdot \vec y\ + \vec y \cdot \vec z\ + \vec z \cdot \vec x$
In terms of the components, what we are trying to find is

$\displaystyle (x_1y_1+x_2y_2+x_3y_3)+(y_1z_1+y_2z_2+y_3z_3)+(z_1 x_1+z_2x_2+z_3x_3)=L$

Rearranging and substituting from the above, we have

$\displaystyle \begin{array}{rcl} 2L&=&(x_1y_1+x_2y_2+x_3y_3)+(z_1x_1+z_2x_2+z_3x_3) \\ &+&(y_1z_1+y_2z_2+y_3z_3)+(z_1x_1+z_2x_2+z_3x_3 )\\ &+&(x_1y_1+x_2y_2+x_3y_3)+(y_1z_1+y_2z_2+y_3z_3 ) \end{array}$

$\displaystyle \Rightarrow\begin{array}{rcl} 2L&=&[(x_1y_1+x_2y_2+x_3y_3)+(z_1x_1+z_2x_2+z_3x_3)]\\ &+&[(y_1z_1+y_2z_2+y_3z_3)+(z_1x_1+z_2x_2+z_3x_3)]\\ &+&[(x_1y_1+x_2y_2+x_3y_3)+(y_1z_1+y_2z_2+y_3z_3)] \end{array}$

$\displaystyle \Rightarrow\begin{array}{rcl} 2L&=&[x_1y_1+z_1x_1+x_2y_2+z_2x_2+x_3y_3+z_3x_3]\\ &+&[y_1z_1+z_1x_1+y_2z_2+z_2x_2+y_3z_3+z_3x_3]\\ &+&[x_1y_1+y_1z_1+x_2y_2+y_2z_2+x_3y_3+y_3z_3] \end{array}$

$\displaystyle \Rightarrow\begin{array}{rcl} 2L&=&[x_1(y_1+z_1)+x_2(y_2+z_2)+x_3(y_3+z_3)]\\ &+&[z_1(x_1+y_1)+z_2(x_2+y_2)+z_3(x_3+y_3)]\\ &+&[y_1(z_1+x_1)+y_2(z_2+x_2)+y_3(z_3+x_3)] \end{array}$

$\displaystyle \Rightarrow\begin{array}{rcl} 2L&=&[x_1(-x_1)+x_2(-x_2)+x_3(-x_3)]\\ &+&[z_1(-z_1)+z_2(-z_2)+z_3(-z_3)]\\ &+&[y_1(-y_1)+y_2(-y_2)+y_3(-y_3)] \end{array}$

$\displaystyle \Rightarrow-2L=\left(x_1^2+x_2^2+x_3^2\right)+\left(z_1^2+z_2^ 2+z_3^2\right)+\left(y_1^2+y_2^2+y_3^2\right)$

$\displaystyle \Rightarrow-2L=4+9+16=29\Rightarrow L=-\frac{29}2$

3. That looks uber complicating but you have the correct answer thanks!
I'm gonna show this to my teacher for further clearification but I bet she won't know rofl.