Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)?explaining it algebraically

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txmedteach | High School Teacher | (Level 3) Associate Educator

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There are two things we need to consider to determine whether this line is in the plane. First, we need to show that the line shares a point with the plane. Then, we need to show that the vector that is perpendicular to the plane is also perpendicular to our line.

Both of these can be most easily determined if we determine the perpendicular for the plane first. To do this, we need to take the cross product of the vectors that are in the parametrization for the plane. In other words, we need to find `vecn` such that:

`vecn = (1,1,-1)xx(2,-1,1)`

Remember, the cross product gives you the vector that is perpendicular to both input vectors. To carry out the cross product, you use the method described in the link below:

In our case, our cross product will look like:

`vecn = veci*det((1, -1),(-1, 1)) - vecj*det((1, -1),(2, 1)) + veck*det((1, 1),(2, -1))`

Simplifying the determinants:

`vecn = veci (1-1) - vecj(1+2) + veck(-1-2)`

This gives us our normal vector:

`vecn = (0, -3, -3)`

So, let's start by determining whether our line is also perpendicular to this normal vector. To find out, let's take the dot product because the dot product of two perpendicular vectors will be 0:

`(1,4,-1)*(0,-3,-3) = 0(1) + 4(-3) + (-3)(-1) = 0 - 12 + 3 = 9`

Our result is clear at this point. Because the vector normal to the plane is not perpendicular to our line, the line cannot be in the plane. Now, we have no reason to proceed to check whether a point is shared by both the line and plane.

It is important that if the line was shown to be perpendicular to the normal vector of the plane that we check to see that at least one point can be shown to be shared between the plane and line. Otherwise, the line and plane may simply never intersect. However, if we show that they do have the same perpendicular vector, and we can show a point is shared, then we know the line must be entirely within the plane. It's a sort of all-or-nothing situation!

I hope that helps!


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