Question

help please
number 11
http://prnt.sc/e6n9y2

Answer 1

@mathmale

Answer 2

For a quick check look at the direction vectors \(\mathbf{AB}\) and \(\mathbf{AC}\)

\(\mathbf{AB} = - \mathbf{OA}\) + \(\mathbf{OB} = \langle 1,3,-4\rangle\)

\(\mathbf{AC} = - \mathbf{OA}\) + \(\mathbf{OC} = \langle -1,-1,1\rangle\)


If \(\mathbf{AB}\) is some multiple of \(\mathbf{AC}\), ie you can solve \(\alpha \mathbf{AB} = \mathbf{AC}\) for some \(\alpha\), then you are moving in the same direction from A to B as you are A to C

You may be moving by a different amount (and perhaps the complete opposite direction, but the direction vectors line up which is the important bit), so they should all be collinear.

This is analogous to using the slope in 2-D, but the slope isn't good enough anymore as you are in \(R^3\).

For a meatier (and better) approach, the equation of the line \(\mathbf{AB}\) is

\(\mathbf l (\lambda) = \mathbf{OA} + \lambda \mathbf{AB}\)

\(= \mathbf{OA} + \lambda (\mathbf{OB} - \mathbf{OA})\)

\(= \langle 2,4,2 \rangle + \lambda \langle 3-2, 7-4, -2-2\rangle\)

\(= \langle 2,4,2 \rangle + \lambda \langle 1, 3, -4 \rangle\)

To verify that B, ie (3,7,-2 ), is indeed on this line, we solve the following for \(\lambda\):

\(\langle 3,7,-2 \rangle = \langle 2,4,2 \rangle + \lambda \langle 1, 3, -4 \rangle\)

We get \(\lambda = 1\)

If C, ie (1,3,3), is also on this line, we can also solve the following for \(\lambda\)

\(\langle 1,3,3 \rangle = \langle 2,4,2 \rangle + \lambda \langle 1, 3, -4 \rangle\)

If we cannot, then A, B, C are not collinear

Answer 3

Thank you ♥