Find an eqquation for The line that passes through the point (3, 43,1) and is parallel to the line F"+3The plane through the points P(2, -1,2) , Q(0,1,2) and ...

Parallel planes Find an equation of the plane parallel to the plane Q passing through the point $P_{0}$ $Q: 4 x+3 y-2 z=12 ; P_{0}(1,-1,3)$

For the given problem, We know that the plane is going to pass through the Points 123. Um And it's gonna be parallel to the xy plane. So if it's parallel to the xy plane, we know that it's normal vector is going to be um It's gonna have zero component, zero component for X and Y. And then a one component for the Z. Direction because it can be parallel to that particular plane. Um So with that in mind that's our normal vector. Um And then that tells us that since we have one times Z minus Z. One in this case the z coordinate is going to be three. So we have one time Z minus three equals zero, which tells us that c minus three equals zero is the equation of our plane that passes through the 00.123 and is also parallel to the X. Y. Plane. That's our final answer.

We need to find the equation of a line that's parallel to the given line and uh passes through the given point. So this is why equals mx plus B. So our slope is three are parallel, slope is also three. So we have our points, we're going to have y minus y. One is equal to m times x minus x one x one, Y one come from the given point. So we have y minus zero is equal to three times X minus zero, Which simplifies to B, Y is equal to three X.

For the given problem, we know that it's parallel to the z axis. So right off the bat, we know our normal vector is going to be um one 10 So with that in mind we can have um with our points since we know that's gonna be the normal vector. And then we're also going to get the vector from the points that we have. We calculate our normal factor and what we end up getting is three 70 So with that we have our normal vector and we have a point given to us. So now we can find the equation which is going to be three X plus seven, Y Equals actually made it 26 equals zero. So that will be the final answer. That's going to be the equation of the plane. We can also move the 26 over here. But in standard form, this is what it should look like.

For the given problem, we want to find the equation Of the plane that passes through the .12, 3 and we know it's parallel to the Y. Z plane. Well, if it's parallel to the Y. Z. Plane and that means that the normal vector at least the direction we know that there is no change in X. Or there's no change in Y. And there's no change in Z. So we have some extraction, no change in y, no change in Z. So we can call this just the unit factory. So we're going to have 100 that's gonna be the direction it goes in. And then um with that in mind we see that that's our normal factor N. We're going to have one times x minus. And then the point we're given is 123 So it can be x minus x. One. In this case, that's gonna be excellence. One equals zero. So our final equation is going to be x minus one equals zero. And again that comes from the fact that if it's parallel to the Y Z plane, then the y and Z components are zero.