Question

Find the scalar equation of the plane that contains the parallel and distinct lines x=1, (y-3)/4=z/2 and x=5, (y+5)/2=(z-3)/1?

Answer 1

Answer is 7x+2y-4z-13=0

Answer 2

the direction vector of the 1º line is(0,4,4) or (0,1,1) to make it more easy
the direction of 2º line is (0,2,1)
To find the plane equation containing this 2 lines find the vector product of this 2 vectors, and later make that the plane pass through one of the points of the lines (1,3,0) or (5,-5,3)

Answer 3

@Ruka

Answer 4

Where did you get those direction vectors?

Answer 5

(y-3)/4=z/4 this is line equation in continuous form

Answer 6

got it?

Answer 7

No.

Answer 8

ok.
P is point, V direction vector. Line equation:
P(a,b,c) + t*V(i,j,k) = (x,y,z)

Answer 9

this is equivalent to 3 equations:
x=a+t*i
y=b+t*j
z=c+t*k
solving all of them for t:
x-a/i =t
y-b/j=t
z-c/k=t
put them all together:
x-a/i=y-b/j=z-c/k
like you can see, in the denominators you have the components of the direction vector and in the nominator the points through which the lione goes

Answer 10

got it?

Answer 11

Where did the other 4 come from?
For the first one, shouldn't it be (0, 4, 2)?
=(0, 2, 1) ?

Answer 12

oh, sry i made a mistake, you right

Answer 13

(0, 4, 2)=(0, 2, 1)

Answer 14

Because the lines are parallel...

Answer 15

you could take perpendicular vector to any of this two
(0,-1,2) and find it's vector product with (0,2,1)

Answer 16

or use the points through which this lines goes to find the other vector, which won't be paralel

Answer 17

Huh?
How do you get a, b, and c?
ax+by+cz+d=0

Answer 18

(5,-5,3) -(1,3,0) =(4,-8,3)

Answer 19

Because the direction vector needs to equal (-7, -2, 4) or (7, 2, -4)

Answer 20

ok.
vector product of (0,2,1) x (4,-8,3)= (14,4,-8)
plane equation using this point (5,-5,3):
14(x-5)+4(y+5)-8(z-3)=0
14x-70 +4y+20-8z+24=0
14x+4y-8z=70-20-24=26
which is you answer multiplied by 2

Answer 21

(5,-5,3) -(1,3,0) =(4,-8,3)

Answer 22

the 2 points of the lines

Answer 23

ok?