Calc: finding an equation

[kid2]

how do i work this?

find an equation for the plane that passes through the point (1,3,1) and contains the line x=t,y=t,z=-2+t

would the line equation be -2i+t(i+j+k)? and then would that mean that the direction or normal vector for the plane be i+j+k? and then what do you do? i have the answer and i can't get it...

 

[stunner5000pt]

to find the equation of a plane you need to have TWO distinct directional vectors and one point the plane passes through

you have been told that it contains the line - what is the directional vector of that line
you need one more directional vector - a vector between the line and the point will give you this. Pick a point on the given line and find the vector between this point and the given point.
the equation of the plane will look like this

$$P = (x,y,z) + sd_{1} + td_{2}$$
where s and t are scalar mutliples, d1 and d2 the directional vectors and x y z is the point

p.s. there are many possible answers

 

[quicknote]

To find the equation of a plane, we need to use the point and normal vector form of the equation, which is (x-x0)⋅n=0, where x0 is a point on the plane and n is the normal vector.

First, we can find the normal vector by taking the cross product of the direction vectors of the given line, which would be (1,1,0) and (0,0,1). This gives us the normal vector (1,-1,1).

Next, we can plug in the given point (1,3,1) and the normal vector into the equation, which gives us (x-1, y-3, z-1)⋅(1,-1,1)=0.

Simplifying this equation, we get x-y+z=3 as the equation of the plane.

In terms of the line equation that you have, it would be helpful to use the parametric form of the line, which is x=1+t, y=3+t, z=-2+t. Then, using the same process as above, we can find the normal vector and ultimately the equation of the plane.

I hope this helps clarify the process for finding the equation of a plane passing through a given point and containing a given line.