How do you find a vector equation and parametric equations in t for the line through the point and perpendicular to the given plane. (P0 corresponds to t = 0.) P0 = (5, 0, 8) x + 2y + z = 9?

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Answer 1 · 2016-06-25T07:16:29

Vector Eqn. :#vecr=<5,0,8>+t*<1,2,1>,# #t in RR.#

Cartesian Eqn. : #x-5=y/2=z-8.#

Observe that the rqd. Line, say, #L# is perp. to the given plane #P : x+2y+z=9.#

So, the direction vector #vecl# of #L# has to be | | to the normal #vecn# of #P#. Here, #vecn = <1,2,1>.#

We choose, #vecl# = #vecn = <1,2,1>.#

Pt. #P_0=(5,0,8) in L.# [Given]

Now, vector eqn. of a line thro. pt. #A# (position vector #veca#) and having dir. along #vecl# is given by, #vecr=veca+t*vecl,# #t in RR.#

Hence, #L : vecr=<5,0,8>+t*<1,2,1>,# #t in RR.#

Its Cartesian Eqn., for #A=(x_1,y_1,z_1), & vecl = (g,f,h)# is given by,

#(x-x_1)/g=(y-y_1)/f=(z-z_1)/h.#

In our case, #L : x-5=y/2=z-8.#