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

#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))#

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

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#.

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

To find the parametric equations of the line of intersection of the planes x + 2z = 0 and 2x - 3y + 4 = 0, we first need to find the direction vector of the line. This direction vector is perpendicular to both normal vectors of the planes. 1. Find the normal vectors of the planes: For the plane x + 2z = 0, the normal vector is <1, 0, 2>. For the plane 2x - 3y + 4 = 0, the normal vector is <2, -3, 0>. 2. Find the direction vector of the line: The direction vector of the line is the cross product of the normal vectors of the planes. <1, 0, 2> × <2, -3, 0> = <0*0 - 2*(-3), 2*0 - 1*0, 1*(-3) - 0*2> = <6, 0, -3> = <2, 0, -1>. 3. Find a point on the line: To find a point on the line, we can set one of the variables (say, z) to a constant (let's use t). From x + 2z = 0, we get x = -2z. Let z = t, then x = -2t. So, a point on the line is (-2t, 0, t). 4. Parametric equations: The parametric equations of the line are: x = -2t, y = 0, z = t.