#1

Describe the intersection between the between the line M: (x-6)/3 = (y-3)/-1 = (z-3)/-2

An the plane Q: x-2y+z+6

My final steps were 3u +3 +d=0

3u+3= -(-6+6-3)

3u =0

u= 0

so Q and M do not intersect or the equation holds for every value of u in which case every point of M lies on Q.
0
4 years ago
#2
(Original post by Vorsah)
An the plane Q: x-2y+z+6
Where is the = ?
0
#3
(Original post by BabyMaths)
Where is the = ?
=0
0
#4
Also can some1 check this:

Find the equation of the plane joining the point x=(0,4,1) y=(1,-2,2) z=(-3,1,-1)

I got 9x-5y+21z+41=0
0
4 years ago
#5
(Original post by Vorsah)
Also can some1 check this:

Find the equation of the plane joining the point x=(0,4,1) y=(1,-2,2) z=(-3,1,-1)

I got 9x-5y+21z+41=0
sub each of the 3 points into the above equation and see if it balances ...

It is quicker than typing the question!
0
#6
is post no1 right then?
0
#7

Q) L= (x-1)=(y-2)=(z-1)/2

M= (x-6)/3 = (y-3)/-1 = (z-3)/-2

find the equation of the plane containing the lines L and M.

first I found the point of intersection of the 2 lines to be (3,4,5).

then I found the equation of the plane to be: 8y-4z-8=0
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#8
can some1 also check this

find the distance between the point X=(0,4,1) and the plane P= 3x+y-z-2=0

I got 1/(sqrt11)
0
4 years ago
#9
(Original post by Vorsah)

Q) L= (x-1)=(y-2)=(z-1)/2

M= (x-6)/3 = (y-3)/-1 = (z-3)/-2

find the equation of the plane containing the lines L and M.

first I found the point of intersection of the 2 lines to be (3,4,5).

then I found the equation of the plane to be: 8y-4z-8=0
Does your point (3,4,5) satisfy 8y-4z-8=0?
0
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