Question

the plane through the point (-1 , 4 , 2) that contains the line of intersection of the planes 4x - y + z -2=0 and 2x +y -2z - 3= 0.

Answer 1

plane thru line of intersection of given planes :

\(4x-y+z-2 + k(2x+y-2z-3) = 0\)

Answer 2

since this is passing thru (-1, 4, 2), plugin and find \(k\) ?

Answer 3

define in detail please

Answer 4

plugin x = -1, y = 4, z = 2 in the above equation and solve \(k\)

Answer 5

we're given that the plane has to go thru that point,
read first line in the question :)

Answer 6

hmmm

Answer 7

on manual solution is different

Answer 8

it doesnt matter what stone we pick to hit our heads

Answer 9

:o

Answer 10

wat is the manual solution ?
post it if u want it explained

Answer 11

ok

Answer 12

Find twopoints P1 and P2 on the line of intersection of the given planes and then find an equation
of the plane that contains P1, P2, and the given point P0(−1, 4, 2). Let (x0,y0,z0)beonthe
line of intersection of the given planes; then 4x0 − y0 + z0 − 2 = 0 and 2x0 + y0 − 2z0 − 3=0,
eliminate y0 by addition of the equations to get 6x0 −z0 −5=0;if x0 = 0 then z0 = −5, if x0 =1
then z0 = 1. Substitution of these values of x0 and z0 into either of the equations of the planes
gives the corresponding values y0 = −7 and y0 =3so P1(0,−7,−5) and P2(1, 3, 1) are on the
line of intersection of the planes.
−→
P0P1 ×
−→
P0P2= 4,−13, 21 is normal to the desired plane whose
equation is 4x − 13y +21z = −14.

Answer 13

you will get the same answer,
solve for \(k\) first

Answer 14

the manual solution looks rather lengthy unnecessarily

Answer 15

ok

Answer 16

k=-8

Answer 17

now what for next??

Answer 18

you should get k = -8/5
http://www.wolframalpha.com/input/?i=4%28-1%29-4%2B2-2+%2B++k*%282%28-1%29%2B4-2%282%29-3%29+%3D+0

Answer 19

plug this k value in the equation :

\(4x-y+z-2 + \frac{-8}{5}(2x+y-2z-3) = 0 \)

simplify, the required plane is :
http://www.wolframalpha.com/input/?i=4x-y%2Bz-2+%2B++%28-8%2F5%29%282x%2By-2z-3%29+%3D+0

Answer 20

ok thnx