Thread: Determine an equation for the plane W that contains the line L1 and is perpendicular

pls help, I can't figure out what to do, I have looked at 2 LA books and neither mention how to work this out. Any help?
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Suppose L1 is the line in R^3 that passes through the points (2, 0, −1) and (3, 1, 1) , and L2 is the line in R^3 defined
by
(x, y, z) = (2, 1, 2) + t (1, 0, 1) , t ∈ R.
Suppose V is the plane in R^3 defined by
−x − y − 2z = 2


A) Determine an equation for the plane W that contains the line L1 and is perpendicular to the plane V?

B) Does the line L2 intersect the plane V? If so, give the point(s) of intersection. If not, give a reason why not?

What ideas have you had so far?

getting the normal to the plane and substitute on of the points of the line into the equation of the plane?

I'm not surprised that you couldn't find this in Linear Algebra books- it is not a Linear Algebra problem. It is a "Calculus III" problem.

You should know that if two planes are perpendicular, then their normal vectors are perependicular. You should also know that if a line lies in a plane, the its "direction vector" is perpendicular to the normal vector to the plane.

So you are given a line and a plane and so can find the direction vector to the line and the normal vector to the plane. The normal vector to the plane you want must be perpendicular to both- use the cross product to find that normal vector.

Once you have the normal vector to the plane, you only need one point in the plane to write its equation. Take any point on the given line.

Originally Posted by HallsofIvy

I'm not surprised that you couldn't find this in Linear Algebra books- it is not a Linear Algebra problem. It is a "Calculus III" problem.

You should know that if two planes are perpendicular, then their normal vectors are perependicular. You should also know that if a line lies in a plane, the its "direction vector" is perpendicular to the normal vector to the plane.

So you are given a line and a plane and so can find the direction vector to the line and the normal vector to the plane. The normal vector to the plane you want must be perpendicular to both- use the cross product to find that normal vector.

Once you have the normal vector to the plane, you only need one point in the plane to write its equation. Take any point on the given line.

thanks, I have a vague idea of what you are saying. I know how to find the cross product of 2 vectors but how do I find the normal\cross product of the plane and the line L1?

so once I have the normal vector (A,B,C), I substitute the values A,B,C into the corresponding equation of the plane Ax + By + Cz + d = 0 and substitute one of the points of L1 ?

Originally Posted by bijosn

thanks, I have a vague idea of what you are saying. I know how to find the cross product of 2 vectors but how do I find the normal\cross product of the plane and the line L1?

You don't. I did not say anything about finding the "cross product of the plane and the line". You don't find the cross product of geometric objects, you find the cross product of vectors. I said "find the cross product of the directional vector of the line and the normal vector of the plane.

so once I have the normal vector (A,B,C), I substitute the values A,B,C into the corresponding equation of the plane Ax + By + Cz + d = 0 and substitute one of the points of L1 ?

Yes.