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

2 Answers
Sep 3, 2016

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.

Sep 3, 2016

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 !