Find the vector equation of the line of intersection for the pair of planes.
Plane one: x+5y-3z-8=0
Plane two: y+2z-4=0
I did half of the work but now i am stuck.
the normal of the planes are not parallel and therefore a solution exists, you simultaneously solve the equations. I got -13z+12=0. Now if i let z = t then how do i find the 'y' and the 'x'
You are almost there.
If you look at the system of equations of the planes
Plane one: x+5y-3z-8=0
Plane two: y+2z-4=0
when converted to the reduced echelon form, you get
x + 0y -13z = -12
0x + y + 2z = 4
Take z as the free variable, (i.e. t) and solve for x and y in terms of t, and z equals t naturally. You will get the solution vector of (x,y,z) in terms of t. Transform this to the appropriate form for your answer.
To find the vector equation of the line of intersection for the pair of planes, you have already correctly solved for z. Since you let z = t, you obtained -13t + 12 = 0.
Now, you can find the values of x and y by substituting the value of z back into the equations of the planes. Let's start with the first plane:
x + 5y - 3z - 8 = 0
Substituting z = t, we get:
x + 5y - 3t - 8 = 0
We can rearrange this equation to solve for x:
x = 3t + 8 - 5y
Now, let's consider the second plane:
y + 2z - 4 = 0
Again, substituting z = t, we have:
y + 2t - 4 = 0
Rearranging this equation to solve for y:
y = 4 - 2t
So, you have found the expressions for x and y in terms of t. The vector equation for the line of intersection can be written as:
r(t) = (3t + 8 - 5y) i + y j + (4 - 2t) k
Alternatively, you can write it as:
r(t) = (3t + 8 - 5(4 - 2t)) i + (4 - 2t) j + t k
where r(t) represents the position vector of any point on the line of intersection and i, j, k are the standard basis vectors in the x, y, and z directions, respectively.