How do you find the parametric equations of the line that passes through #A=(7,-2,4)# and that is parallel to the line of intersection of the planes #4x-3y-z-1=0# and #2x+4y+z-5=0#?

2 Answers
Cesareo R.
Sep 27, 2016

#{(x=7+gamma),(y=-2-6gamma),(z = 4+22gamma):}#

Explanation:

The plane can be represented by

#Pi-> << p - p_0, vec n >> = 0# or

#(x-x_0)a+(y-y_0)b+(z-z_0)c=0# or
#ax + by + cz -ax_0-by_0-cz_0=0#

Here #p = (x,y,z)#, #p_0 = (x_0,y_0,y_0)# and #vec n = (a,b,c)#

So

#Pi_1-> << p-p_1, vec n_1 >> = 0# and
#Pi_2-> << p-p_2, vec n_2 >> = 0#

with

#p_1 = (0,0,-1)# and #vec n_1 = (4,-3,-1)#
#p_2 = (0,0,5)# and #vec n_2 = (2,4,1)#

Let

#L_i->p = p_i+lambda vec v# the intersection line between the two planes

If #L_i = Pi_1 nn Pi_2# then

#<< p_i-p_1+lambda vec v, vec n_1 >> = 0# and
#<< p_i-p_2+lambda vec v, vec n_2 >> = 0#

or

#<< p_i-p_1, vec n_1 >> + lambda << vec v, vec n_1 >> = 0#
#<< p_i-p_1, vec n_2 >> + lambda << vec v, vec n_2 >> = 0#

but #p_i in Pi_1 nn Pi_2# so

#<< p_i-p_1, vec n_1 >> = 0# and
#<< p_i-p_1, vec n_2 >>=0#

then

#vec v = mu vec n_1 xx vec n_2# because it must be orthogonal to both.

So, the line parallel to #L_i# passing by #A# is

#L_A->A+gamma vec n_1 xx vec n_2#

#n_1 xx n_2 = (1,-6,22)# so the parametric equations are

#{(x=7+gamma),(y=-2-6gamma),(z = 4+22gamma):}#

Ratnaker Mehta
Sep 27, 2016

The reqd. eqn. of line :
#vecr=(x,y,z)=(7,-2,4)+t(1,-6,22), t in RR.#

Explanation:

Look at my Solution, based upon more of Algebra , and a little bit of

Geometry!

Let the given planes be

#pi_1 : 4x-3y-z-1=0...(i), &, pi_2 : 2x+4y+z-5=0...(ii)#

#(i)+(ii) rArr 6x+y-6=0, i.e., y=6-6x... ... ... (1)#

Sub.ing in #(i)#,

#4x-3(6-6x)-z-1=0, i.e., 22x-19=z... ... ...(2)#

Altogether, we have,

#y=6-6x, z=22x-19, 7", of course, "x=x#

Since the above have been derived by solving #pi_1 & pi_2#, this

means that, #AA x in RR, (x,6-6x,22x-19) in pi_1nnpi_2, or,#

#pi_1nnpi_2={(0,6,-19)+x(1,-6,22) : x in RR}#

Observe that, since the reqd. line is #||# to #pi_1nnpi_2# the

direction of the reqd. line is (along) the vector #(1,-6,22)#

Finally, using the pt. #A(7,-2,4)# on the reqd. line, we get, its eqn.

#vecr=(x,y,z)=(7,-2,4)+t(1,-6,22), t in RR#

Enjoy Maths.!

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