Vector

[123]

HELLO,

\(\displaystyle ABCDA_1B_1C_1D_1\) - cube. Find the cosine angle between vectors \(\displaystyle \vec{AC}\) and \(\displaystyle \vec{AC_1}\).

 

[royhaas]

Draw a picture and consider the dot product.

 

[123]

 

[123]

Help me

 

[mmm4444bot]

Help me

I'm guessing that the instruction is supposed to read "Find the cosine of the angle…".

It's hard to help you, without knowing what you've already learned about vector arithmetic or where you're stuck.

What have you done so far? Do you have a specific question? :?

 

[soroban]

Hello, 123!

\(\displaystyle \text{Cube: }\,ABCDA_1B_1C_1D_1\)

\(\displaystyle \text{Find the cosine of the angle between vectors }\overrightarrow{AC}\text{ and }\overrightarrow{AC_1}\)

You drew a very nice diagram.
I assume you know the Angle formula.


\(\displaystyle \text{Place }A\text{ at the origin.}\)
. . \(\displaystyle \text{Then: }\:A(0,0,0),\;B(1,0,0),\;C(1,1,0),\;D(0,1,0),\;A_1(0.0.1),\;B_1(1,0,1),\;C_1(1,1,1),\;D_1(0,1,1)\)

\(\displaystyle \text{Then: }\:\overrightarrow{AC} \,=\,\langle 1,1,0\rangle,\;\overrightarrow{AC_1} \,=\,\langle 1,1,1\rangle\)

\(\displaystyle \text{Formula: }\;\cos\theta \;=\;\dfrac{\overrightarrow{AC}\cdot\overrightarrow{AC_1}}{|\overrightarrow{AC}|\,|\overrightarrow{AC_1}|} \;=\;\dfrac{\langle 1,1,0\rangle\cdot\langle1,1,1\rangle}{\sqrt{1^2+1^2+0^2}\,\sqrt{1^2+1^2+1^2}}\)

\(\displaystyle \text{Can you finish it?}\)

 

[123]

\(\displaystyle }\;\cos\theta \;=\;\dfrac{\overrightarrow{AC}\cdot\overrightarrow{AC_1}}{|\overrightarrow{AC}|\,|\overrightarrow{AC_1}|} \;=\;\dfrac{\langle 1,1,0\rangle\cdot\langle1,1,1\rangle}{\sqrt{1^2+1^2+0^2}\,\sqrt{1^2+1^2+1^2}}=\frac{\sqrt{6}}{3}\)