coplanar vectors

[justsomeguy]

the question is:

The vectors A, B and C are given by: A = −2i + 5j − 6k, B = 2i + 4j + 5k and C = 8i − j − 2k. Decide whether or not these points are co-planar.

I know I need to use A · B × C = 0, I tried it but just ended up with such large numbers it felt wrong, so wanted to ask for help.

 

[Deleted member 4993]

the question is:

The vectors A, B and C are given by: A = −2i + 5j − 6k, B = 2i + 4j + 5k and C = 8i − j − 2k. Decide whether or not these points are co-planar.

I know I need to use A · B × C = 0, I tried it but just ended up with such large numbers it felt wrong, so wanted to ask for help.

You have the correct idea.

Please share your work - with all the steps - so that we can catch your "mis-step"!

 

[HallsofIvy]

\(\displaystyle B\times C= \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 & 4 & 5 \\ 8 & -1 & -2\end{array}\right|= \vec{i}\left|\begin{array}{cc} 4 & 5 \\ -1 & -2 \end{array}\right|- \vec{j}\left|\begin{array}{cc}2 & 5 \\ 8 & -2\end{array}\right|+ \vec{k}\left|\begin{array}{cc} 2 & 4 \\ 8 & -1 \end{array}\right|\)
\(\displaystyle = -3\vec{i}+ 44\vec{j}- 34\vec{k}\).
You think "44" and "-34" are "such large numbers"?

Then \(\displaystyle A\cdot B\times C= (-2)(-3)+ (5)(44)+ (-6)(-34)= 6+ 220+ 204= 430\).

Nothing wrong with that!

 

[justsomeguy]

\(\displaystyle B\times C= \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 2 & 4 & 5 \\ 8 & -1 & -2\end{array}\right|= \vec{i}\left|\begin{array}{cc} 4 & 5 \\ -1 & -2 \end{array}\right|- \vec{j}\left|\begin{array}{cc}2 & 5 \\ 8 & -2\end{array}\right|+ \vec{k}\left|\begin{array}{cc} 2 & 4 \\ 8 & -1 \end{array}\right|\)
\(\displaystyle = -3\vec{i}+ 44\vec{j}- 34\vec{k}\).
You think "44" and "-34" are "such large numbers"?

Then \(\displaystyle A\cdot B\times C= (-2)(-3)+ (5)(44)+ (-6)(-34)= 6+ 220+ 204= 430\).

Nothing wrong with that!

Thank you, that's exactly what I got! Just they were larger than the numbers I would always use when I first learned vectors. Thanks.