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

**narutoblaze** · February 17th 2009, 03:43 PM

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 $\vec x \cdot \vec y\ + \vec y \cdot \vec z\ + \vec z \cdot \vec x$ , 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

**Reckoner** · February 17th 2009, 04:48 PM

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:

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

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

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

Three vectors x, y, and z satisfy x + y + z = 0.

This tells us that

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

|x| = 2, |y| = 3, and |z| = 4

Using the dot product property $\textbf u\cdot\textbf u=\lVert\textbf u\rVert^2,$ we have

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

Calculate the value of $\vec x \cdot \vec y\ + \vec y \cdot \vec z\ + \vec z \cdot \vec x$

In terms of the components, what we are trying to find is

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

Rearranging and substituting from the above, we have

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

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

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

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

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

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

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

**narutoblaze** · February 17th 2009, 05:29 PM

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.

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