# scalar triple product, coplanar vectors.

• Feb 23rd 2009, 11:54 PM
rcmango
scalar triple product, coplanar vectors.
a = 2i+ 3j +k,
b = i - j,
c = 7i + 3j +2k

must use the scalar triple product to conclude that they are coplanar.

i tried this using a *(b x c)

i also said b was b = i - j +0k ??

so i came out with

2(-2-0) + 3(2-0) + 1(3 - -7)

for -4 +6 + 10

don't think this is correct though, i think they need to be 0.

thanks for any help.
• Feb 24th 2009, 04:47 AM
Soroban
Hello, rcmango!

You dropped a minus-sign . . .

Quote:

$\displaystyle \begin{array}{c}a \:=\:2i+ 3j +k \\ b \:=\: i - j \\ c \:=\: 7i + 3j +2k\end{array}$

Use the scalar triple product to conclude that they are coplanar.

We have: . $\displaystyle \begin{array}{c}a \;=\:\langle2,3,1\rangle \\ b \:=\:\langle1,\text{-}1,0\rangle \\ c \:=\:\langle 7,3,2\rangle \end{array}$

. . . . . . . . .$\displaystyle {\color{red}^{+\quad-\quad+} }$
$\displaystyle a\cdot(b\times c) \;=\; \begin{vmatrix}2&3&1 \\ 1&\text{-}1&0 \\ 7&3&2\end{vmatrix}\;=\;2(\text{-}2-0) \:{\color{red}-}\:3(2-0) \:+\: 1(3-[\text{-}7])$

• Feb 24th 2009, 10:24 AM
rcmango
thanks, ya i'm finding out its really easy to miss signs in these types of problems.