What is parametric equation of the line created by the intersecting planes x = 2 and z = 2?

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Answer 1 · 2016-08-02T12:45:33

#vec r = ((2),(0),(2)) + lambda ((0),(1),(0))#

no need to do much here because every point on the line of intersection will have values x = 2 and z = 2, and any value of y is possible.

a fixed point on the line is (2,0,2) so we can say that the line is this....

#vec r(lambda) = ((2),(0),(2)) + lambda ((0),(1),(0))#

Answer 2 · 2016-08-03T17:05:24

#vecr=(2,0,2)+t(0,1,0), t in RR#, or,

in the usual Cartesian Form # (x-2)/0=y/1=(z-2)/0#.

Let the given planes be #pi_1 : x-2=0, and pi_2 : z-2=0#.

Let the reqd. line be #L=pi_1 nn pi_2#.

To determine the eqn. of #L#, we need, #(1)# a pt., say, #A in L#, &

#(2)# the direction #vecl# of #L#.

(1) : The point A :-.

#A(x,y,z) in L=pi_1nnpi_2rArr A in pi_1, and, A in pi_2#

#rArr x=2, z=2#. As regards, #y in RR#, y is arbitrary, and, by our

choice, #y=0#. Hence, #A=A(2,0,2)#

(2) : The Direction #vecl# of #L# :-

Let #vecn_1, &, vecn_2# be the normals to #pi_1, &, pi_2#, resp.

#:. vecn_1=(1,0,0)=hati, &, vecn_2=(0,0,1)=hatk#

#L=pi_1nnpi_2 rArr L sub pi_1, &, L sub pi_2 rArr vecl bot vecn_1, &, vecl bot vecn_2#.

#rArr vecl# is along #vecn_1xxvecn_2=hatixxhatk=-hatj=(0,-1,0)#.

we take, #vecl=hatj=(0,1,0)#

Using #(1), &, (2)#, the vector eqn., or, parametric eqn. of #L :#

#vecr=veca+tvecl, t in RR#, where, #veca# is the position vector of

the pt.#A#.

Hence, # L : vecr=(2,0,2)+t(0,1,0), t in RR#, or, in the usual

Cartesian Form # L : (x-2)/0=y/1=(z-2)/0#.

Hope, this will be helpful! Enjoy Maths!