# points in a plane

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• Apr 6th 2011, 02:32 AM
mechaniac
points in a plane
determine the constant $\beta$ so that the points $A=(0,-8,-3),B=(3,\beta,1),C=(-1,0,0),D=(2,1,1)$ is in a plane.

I started to put up two vectors with the points A,B,C:

$\bar{AB}=(3,\beta+8,4)$

$\bar{AC}=(-1,8,3)$

now i thought that if i choose $\beta$ so that the two vectors makes a plane, i can get the equation for the plane and proof that the point D is in the plane.

How do i choose $\beta$ so the vectors makes a plane?

Thanks!
• Apr 6th 2011, 02:59 AM
Plato
Quote:

Originally Posted by mechaniac
determine the constant $\beta$ so that the points $A=(0,-8,-3),B=(3,\beta,1),C=(-1,0,0),D=(2,1,1)$ is in a plane. How do i choose $\beta$ so the vectors makes a plane?

Can you solve this $\overrightarrow {AB} \cdot \left( {\overrightarrow {AC} \times \overrightarrow {AD} } \right) = 0~?$
• Apr 6th 2011, 03:40 AM
mechaniac
yes but isnt it only in space $(R^{3})$, $(\bar{AC}\times\bar{AD})$ works?
• Apr 6th 2011, 03:53 AM
Deveno
your vectors ARE in R^3, yes?
• Apr 6th 2011, 04:08 AM
mechaniac
oh sorry got confused by the words. thought they want it in R^2 when they say plane and r^3 when it says space :P
• Apr 6th 2011, 06:55 AM
HallsofIvy
Quote:

Originally Posted by mechaniac
determine the constant $\beta$ so that the points $A=(0,-8,-3),B=(3,\beta,1),C=(-1,0,0),D=(2,1,1)$ is in a plane.

I started to put up two vectors with the points A,B,C:

$\bar{AB}=(3,\beta+8,4)$

$\bar{AC}=(-1,8,3)$

now i thought that if i choose $\beta$ so that the two vectors makes a plane, i can get the equation for the plane and proof that the point D is in the plane.

How do i choose $\beta$ so the vectors makes a plane?

Thanks!

Any two (independent) vectors will define a plane. You can write the plane in the form $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= D$ where $(x_0, y_0, z_0)$ is a point in the plane (so any of the points you are given) and <A, B, C> is the cross product of the two vectors in the plane.

So, you could, in fact, write the equation of the line, with $\beta$ in the coefficients, then put the coordinates of each point and find what value of $\beta$ is required so that each point satisfies the equation of the plane.

However, simpler is what Plato says- the cross product of two vectors in the plane is perpendicular to the plane and so has 0 dot product with any vector in the plane. That is, you can take the cross product of any two of the vectors, say AC ahd AD and then take the dot product of that with the third, AB. The four points will lie in one plane if and only if that is 0.