a vector parallel to the line of intersection of the planes. x-2y-z=6 and 3x-y+z=4

the answer is <3,4,-5>

so first, I took the cross product of <1,-2,-1> and <3,-1,1> which I think is v. in the equation p0 + tv. << this is the equation for the line of intersection I am pretty sure. But, I'm not sure what to do after that.

+i +j +k

+1 -2 -1
+3 -1 +1
================== cross product
(-3)i + (-4)j + (+5)k

so yes V = -3 i - 4 j + 5 k
opposite to that is also parallel
+3 i + 4 j -5 k

add any old point to that and it is still parallel

another way: find the intersection line of the two planes

add the two equations,
4x -3y = 10
y = (4x-10)/3

pick any "nice" value of x

let x = 1, y = -2
in the 2nd equation:
3 + 2 + z = 4 , z = --1 --> point (1,-2,-1)

let x = 10, y = 10,
in the 2nd
30 - 10 + z = 4, z = -16 -->point(10, 10, -16)

vector = <9,12,-15> or reduced to <3, 4, -5>

To find a vector parallel to the line of intersection of the two planes, you correctly took the cross-product of the normal vectors of the planes. The cross product of <1,-2,-1> and <3,-1,1> is:

<1,-2,-1> × <3,-1,1> = [(-2)(1) - (-1)(-1), (-1)(3) - (1)(1), (1)(3) - (-1)(-2)]
= [-1 + 1, -3 - 1, 3 + 2]
= [0, -4, 5]

So, the vector v = <0, -4, 5> is parallel to the line of intersection of the planes.

Now, we need to find a point on the line of intersection, which we can call p0. We can choose any valid point that lies on both planes. So, let's solve the system of equations for x, y, and z:

x - 2y - z = 6 (Equation 1)
3x - y + z = 4 (Equation 2)

Multiplying Equation 1 by 3, we get:

3x - 6y - 3z = 18 (Equation 3)

Adding Equation 3 to Equation 2, we eliminate z:

(3x - y + z) + (3x - 6y - 3z) = 4 + 18
6x - 7y = 22 (Equation 4)

Now, let's solve Equation 4 for x, in terms of y:

6x = 7y + 22
x = (7y + 22)/6

Substituting this value of x into Equation 1, we can get a relationship between y and z:

(7y + 22)/6 - 2y - z = 6
7y + 22 - 12y - 6z = 36
-5y - 6z = 14 (Equation 5)

Let's solve Equation 5 for y, in terms of z:

-5y = 14 + 6z
y = (-14 - 6z)/5

Now, substituting the values of y and x into Equation 1 or Equation 2, we can solve for z. Let's substitute the values into Equation 1:

x - 2y - z = 6
(7y + 22)/6 - 2y - z = 6
(7(-14 - 6z)/6 + 22)/6 - 2(-14 - 6z)/6 - z = 6
-98/6 - 7z/3 + 22/6 + 28/3 - z = 6
-49/3 - 7z/3 + 22/6 + 56/6 - z = 6
-49/3 - 14z/6 + 22/6 + 56/6 - 6z/6 = 6
(-98 - 14z + 22 + 56 - 6z)/6 = 6
(-8z - 20)/6 = 6
-8z - 20 = 36
-8z = 56
z = -7

Now, substituting the value of z back into Equation 5, we can solve for y:

-5y - 6(-7) = 14
-5y + 42 = 14
-5y = -28
y = 28/5 = 5.6

Finally, substitute the values of y and z back into Equation 4 to solve for x:

6x - 7(5.6) = 22
6x - 39.2 = 22
6x = 22 + 39.2
6x = 61.2
x = 10.2

So, the point p0 on the line of intersection is (10.2, 5.6, -7).

Therefore, the equation of the line of intersection is given by p0 + tv, where p0 = (10.2, 5.6, -7) and v = <0, -4, 5>.

To find a vector parallel to the line of intersection of the planes, we can use the cross product of the normal vectors of the two planes. Here's how you can proceed:

Step 1: Write the equations of the planes in the form "Ax + By + Cz = D".
For the first plane, x - 2y - z = 6, we can assign A = 1, B = -2, C = -1, and D = 6.
For the second plane, 3x - y + z = 4, we can assign A = 3, B = -1, C = 1, and D = 4.

Step 2: Find the normal vectors of the planes.
The normal vector of the first plane can be found by taking coefficients of x, y, and z: n1 = <A, B, C> = <1, -2, -1>.
The normal vector of the second plane can be found in the same way: n2 = <3, -1, 1>.

Step 3: Calculate the cross product of the normal vectors.
The cross product of n1 and n2 will give us a vector that is orthogonal to both normals and therefore parallel to the line of intersection. We can calculate this using the cross product formula:
v = n1 x n2
v = <1, -2, -1> x <3, -1, 1>

To find v, we can use the following determinant formula:

| i j k |
| 1 -2 -1 |
| 3 -1 1 |

Expanding this determinant, we get:

v = (1 * (-1) - (-2) * (-3))i - ((1 * 1) - (-2) * 3)j + (1 * (-1) - (-2) * 3)k
= (-1 + 6)i - (1 - 6)j + (-1 + 6)k
= 5i - (-5)j + 5k
= 5i + 5j + 5k
= 5(1, 1, 1)
= <5, 5, 5>

Step 4: Normalize the vector if desired.
Since we want a vector parallel to the line of intersection, we don't need to normalize it. However, if you prefer a unit vector, you can divide v by its magnitude to get the unit vector parallel to the line.

Therefore, a vector parallel to the line of intersection is <5, 5, 5>.