Question

How do you find the parametric equations of the line of intersection of the planes x+2z=0 and 2x-3y+4?

Answer 1

`vec l = ((0),(4/3),(0)) + t ((6),(4),(-3))` where `t` is the parameter.

Explanation:

`pi_1: x + 2z = 0`
`pi_2: 2x -3y =-4`

the line will lie on both planes, clearly. As such the equation of each plane in form `pi_(i): (vec r_i - vec r_(o \ i) )* vec n_i = 0` will also hold true for every point on the line apropos both planes.

So the line `vec l = vec l_o + vec d` will run in a direction `vec d` that is perpendicular to the `vec n_(1,2)` and its direction will be `vec d = vec n_1 times vec n_2`

ie `vec d = det ((hat x, hat y, hat z),(1,0,2),(2,-3,0)) hat n_d`

`=6 hat x + 4 hat y -3 hat z = ((6),(4),(-3))`

All we now need is a point on the line. if we set x = 0 for both `pi_1` and `pi_2` then we see that `z = 0, y = 4/3` or `vec l_o = ((0),(4/3),(0))`

therefore:

`vec l = ((0),(4/3),(0)) + t ((6),(4),(-3))` where `t` is the parameter.

Answer 2

`x/6=(y-4/3)/4=z/-3` is the reqd. Cartesian Eqn.,

OR,

`vecr=(0,4/3,0)+lambda(6,4,-3), lambda in RR`

Explanation:

Let us denote, by `pi_1 and pi_2` the given planes, resp.

From `pi_1, x=-2z.................................................(1)`

Sub.ing this `x` in `pi_2,` we get,

`-4z-3y+4=0, or, 3y-4=-4z..................(2)`

Rewriting `(1)` as, `2x=-4z.....................................(1')`

And, Obviously, `-4z=-4z......................................(3)`

Combining `(1'), (2), and (3)`, we have,

`2x=3y-4=-4z, i.e., 2x=3(y-4/3)=-4z`, or,

`(2x)/12=(3(y-4/3))/12=(-4z)/12`,

`:. x/6=(y-4/3)/4=z/-3` is the reqd. Cartesian Eqn. of

the Line `sub pi_1nnpi_2`.

Its vector eqn. can be written as

`vecr=(0,4/3,0)+lambda(6,4,-3), lambda in RR`, in accordance with

the Answer submitted by Respected Eddie !