Example of Symmetric Equations of a Line

Multivariable Calculus: Find the symmetric equations of the line through the point (1,0,3) and perpendicular to the plane x+2y-z=6.

For more videos like this one, please visit the Multivariable Calculus playlist at my channel.

Category:

Education

Tags:

• Mathematics
• multivariable
• calculus
• several
• variables
• equation
• line
• symmetric
• plane
• perpendicular

[zacnoob]

I got a question i not very sure of:

Conside the two planes with equations: 3x+4y=1 & x+y-z=3, determine the symmetric equations for the line of intersection L of these two planes?

[MathDoctorBob]

@zacnoob I have a video for that: Line of Intersection of 2 Planes

Youtube address: jozabh0lFmo

The short version is that the direction of the line is the cross product of the normal vectors. Then you need to solve the plane equations together to find a point in both planes. Let me know if that helps. I'm happy to answer any questions you might have. - Bob

[zacnoob]

@MathDoctorBob But one of the equation only have x & y axis while the other got three, so if i put z=0, the equation in the end will still be not affected for this question?

[MathDoctorBob]

@zacnoob Yes, the normal vectors are (3,4,0) and (1,1,-1). It helps to think of 3x+4y=1 as 3x+4y+0z=1. Convenient notation confuses the problem. - bob

[MathDoctorBob]

@guitarguitarguitar10 OK. Different problem. We're want a plane, not a line. The line joining (16,0,0) and (0,0,16) is in the plane and has direction (1,0,-1). The normal direction of the given plane is (4,-1,2). The normal direction of the new plane is given by the cross product of (1,0,-1) and (4,-1,2) = (-1,-6,-1). Of course, you should draw the picture to see how things work.

So the new plane has equation -x-6y-z=d. To get d, plug in either point to get d = -16. - Bob

[guitarguitarguitar10]

@MathDoctorBob But how would I solve it if it was perpendicular and had two points? I have a problem with two points again (0,0,16) and (16,0,0) with the plane 4x-y+2z=7. The answer is -x-6y-z+16=0, but i"m not sure how calculate it.

[MathDoctorBob]

@guitarguitarguitar10 You're welcome!

It takes two points to form a line, so we have "too much" information. This means there may be no solution. There could be a solution - we find it from the points and check if it is perpendicular to the plane.

The line through the two points has direction (3, 0, -4). The normal direction of the plane is (3,2,6). Since these are not multiples of one another, the line is not perpendicular to the plane. So no solution. - Bob

[guitarguitarguitar10]

Hey I had a quick question. How would you solve it if you had two points?

In the problem I have to do the points are (1,2,3) and (4,2,-1) and the plane is 3x +2y+6z=0. Thanks your videos are really helpful!!!

[MathDoctorBob]

@gyg0312 Thanks for the comment.  Unintentional, but I've done worse. :)

[gyg0312]

so helpful.. and haha se:x-1 xD