How do you determine if the vectors are coplanar: a = [-2,-1,4], b = [5,-2,5], and c = [3,0,-1]?

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Answer 1 · 2016-07-28T18:14:38

They are coplanar.

Evaluate the determinant of the matrix with these vectors as rows:

$abs((-2, -1, 4),(5,-2,5),(3,0,-1))$

$=-2 abs((-2,5),(0,-1)) -1 abs((5,5),(-1,3))+4 abs((5,-2),(3,0))$

$=-2(2)-1(20)+4(6)$

$=-4-20+24$

$=0$

Since the determinant is $0$ , these vectors only span at most a $2$ dimensional space. That is, they are coplanar.

Answer 2 · 2016-07-28T20:12:13

See below

Two non linearly dependent vectors conveniently parametrized, plus a point, generate a plane

$Pi-> p_0 +lambda_1 vec a + lambda_2 vec b$ .

with the plane tangent space given by.

$t_{Pi}->lambda_1 vec a + lambda_2 vec b$ with ${lambda_1,lambda_2} in RR^2$

Now if $vec c in t_{Pi}$ then $EE {lambda_1,lambda_2} in RR^2 | lambda_1 vec a + lambda_2 vec b = vec c$

Solving for $lambda_1,lambda_2$ we have

${ (-2 lambda_1 + 5 lambda_2 = 3), (-lambda_1 - 2 lambda_2 = 0), (4 lambda_1 + 5 lambda_2 = -1) :}$

$lambda_1 = -2/3,lambda_2 = 1/3$

so $vec c in t_{Pi}$