Equations of 2 intersecting planes are given. Find 2 pts w/ integral coordinates on that line of intersection?

Question

The equation of 2 intersecting planes are 4x-3y+6z=41 and x+2y-3z=5. Determine 2 pts w/ integral coordinates that are on their line of intersection

so where do i start to solve this problem?

here's the problem. using the method of guess and check is easy for me to solve this problem but i want to know how to solve this problem w/o using guess and check

Answer 1

solve each equation for, say z in terms of x and y, then set the two expressions equal. in this case, we'll solve for 6z instead of z( makes computations easier)

6z = 41 - 4x + 3y

6z = -10 +2x + 4y

so -10 + 2x + 4y = 41- 4x +3 y

thus y = -6x + 51.

6z = -10 +2x +4(-6x + 51)
6z = 194 -22x

need values of x which make 194-22x a multiple of 6 (because this equals 6z), for example x=-2, x=1. solving for y and z with these values of x we get (-1,57,36) and (2, 39, 25)

Finally, we check that both points satisfy the original equations:

4(-1) - 3(57) + 6(36) = -4 -171 + 216 = 41;
(-1) + 2(57) - 3(36) = -1 + 114 - 108 = 5;

so the first point is is on the intersection. Now, the second:

4(2) - 3(39) +6(25) = 8 - 117 +150 = 41
(2) + 2(39) - 3(25) = 2 + 78 - 75 = 5

so the second point is also on the intersection. We could find more points, corresponding to x = 5,8,11, etc. and x=-4,-7,-10, etc. (can you see why these x's will give integer values for y and z? it's sort of hard: basically, 194 - 22x has to be a multiple of 6. This happens for the x-values described).

Answer 2

You can also convert the line equation to y^2=(x-4)^2 and then you have 2x^2 - 17 = (x-4)^2 For solving: 2x^2 -17 = x^2-8x+16 x^2 + 8x - 33 = 0 (x + 11) (x - 3) = 0 The solutions of the equation are two for x : -11 and 3. Just substitute those values in the line equation to get the values of y. y1= x1-4= -11 - 4 = -15 y2= x2-4 = 3-4 = -1 So the coordinates are (-11, -15) and ( 3, -1)

Answer 3

Find the line of intersection of the two given planes.

4x - 3y + 6z = 41
x + 2y - 3z = 5

The cross product of the normal vectors of the two planes will be normal to both of them and therefore the directional vector of the line of intersection.

<4,-3,6> X <1,2,-3> = -3i + 18j + 11k

Now find a point on the line.

4x - 3y + 6z = 41
x + 2y - 3z = 5

Multiply the second equation by 4 and we have:

4x - 3y + 6z = 41
4x + 8y - 12z = 20

Solve for 4x for each equation and set equal to each other.

4x = 3y - 6z + 41
4x = -8y + 12z + 20

3y - 6z + 41 = -8y + 12z + 20
11y = 18z - 21
y = (18z - 21)/11

Let z = 3.
Then y = (18*3 - 21)/11 = (54 - 21)/11 = 33/11 = 3

Now plug into one of the equations and solve for x.

x + 2y - 3z = 5
x = -2y + 3z + 5
x = -2*3 + 3*3 + 5 = 8

So we have the point (8,3,3) on the line of intersection.

To find another point take the equation for the line:

L: <8,3,3> + t<-3,18,11>

Set t = 1 and get another point

(8-3,3+18,3+11) = (5,21,14)

So two points on the line with integral coordinates are:

(8,3,3) and (5,21,14)

Answer 4

substitute z
then you will obtain the line equation

then chose 2 arbitrary values for x then solve for y

in this way you will obtain the coordinates for 2 points