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Why is this theorem on coplanar vectors true (LINEAR ALGEBRA)? 
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#1
Feb2512, 05:19 PM

P: 11

A necessary and sufficient condition for three vectors to be coplanar is the equality is that the determinant of the matrix equals zero.



#2
Feb2512, 05:43 PM

Sci Advisor
P: 2,684

Basically that just says that [itex]A\cdot (B\times C)=0[/itex] (or any reordering thereof). BXC is a vector that is perpendicular to B and C. If A is coplanar with B and C, then it can be expressed as a linear combination of the two, i.e. A=bB+cC where b and c are real numbers. In that case, then it's obvious that A dotted into this vector which is perpendicular to both B and C would be 0.
Another way to think about it is to note that the above triple product has a value which is the volumn of the parallelepiped defined by A, B and C. If A, B and C are coplanar, then the parallelepiped has 0 volume. 


#3
Mar112, 04:44 AM

P: 233

What Matterwave said is correct it follows from a theorem that says the determinant of a matrix is non zero if and only if the vectors which make it up are all linearly independent. So if the vectors are coplanar obviously one the vectors is linearly dependent so the determinant of the matrix they form must be zero.



#4
Mar112, 04:52 AM

Sci Advisor
P: 906

Why is this theorem on coplanar vectors true (LINEAR ALGEBRA)?
do this, find this determinant:
[tex]\begin{vmatrix}x_1&y_1&ax_1+by_1\\x_2&y_2&ax_2+by_2\\x_3&y_3&ax_3+by_3 \end{vmatrix}[/tex] the results should be enlightening. 


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