Find the intersection point of the lines {p:p = (-2, 4, 6) + t(1, 2, 3)} and {p:p = (-2, 1, 6) + t(3, 2, 1)}. Find equation of the plane containing the two lines
To find the intersection point of the lines, we set the two equations equal to each other:
(-2, 4, 6) + t(1, 2, 3) = (-2, 1, 6) + s(3, 2, 1)
Simplifying, we get:
(-2 + t, 4 + 2t, 6 + 3t) = (-2 + 3s, 1 + 2s, 6 + s)
Equating the x-components, we have:
-2 + t = -2 + 3s
Simplifying, we get:
t - 3s = 0
Equating the y-components, we have:
4 + 2t = 1 + 2s
Simplifying, we get:
2t - 2s = -3
Equating the z-components, we have:
6 + 3t = 6 + s
Simplifying, we get:
3t - s = 0
We now have a system of equations that we can solve to find the values of t and s. Rewriting the first equation, we get t = 3s. Substituting this into the second equation, we get 2(3s) - 2s = -3, which simplifies to 4s = -3. Solving this equation gives us s = -3/4. Substituting this back into the first equation, we get t = 3(-3/4), which simplifies to t = -9/4.
Therefore, the intersection point of the lines is:
(-2 + (-9/4), 4 + 2(-9/4), 6 + 3(-9/4)) = (-17/4, -1/2, 15/4)
To find the equation of the plane containing the two lines, we need to find two vectors that are parallel to the lines. We can take the direction vectors of the lines as these two vectors. The direction vector of the first line is (1, 2, 3), and the direction vector of the second line is (3, 2, 1).
Now we can find the normal vector of the plane by taking the cross product of these two vectors:
(1, 2, 3) × (3, 2, 1) = (4, -8, 4)
The equation of the plane is then of the form:
ax + by + cz + d = 0
Substituting the coordinates of the intersection point into this equation, we get:
(4)(-17/4) + (-8)(-1/2) + (4)(15/4) + d = 0
Simplifying, we get:
-17 + 4 + 15 + d = 0
Solving for d, we get d = -2.
Therefore, the equation of the plane containing the two lines is:
4x - 8y + 4z - 2 = 0.
To find the intersection point of the two lines, we can set their parametric equations equal to each other and solve for t.
For the first line:
p = (-2, 4, 6) + t(1, 2, 3)
For the second line:
p = (-2, 1, 6) + t(3, 2, 1)
Setting them equal to each other:
(-2, 4, 6) + t(1, 2, 3) = (-2, 1, 6) + s(3, 2, 1)
Now we can equate the x, y, and z components separately:
-2 + t = -2 + 3s
4 + 2t = 1 + 2s
6 + 3t = 6 + s
We can solve these equations to find the values of t and s. Subtracting the second equation from the first gives:
3 + t - 2t = s
3 = s - t
Substituting this value of s into the other two equations:
4 + 2t = 1 + 2(3 - t)
6 + 3t = 6 + (3 - t)
Simplifying these equations:
4 + 2t = 1 + 6 - 2t
6 + 3t = 6 + 3 - t
Combining like terms:
4t = 3 + 2
4t = 5
t = 5/4
Substituting this value of t into the equation s = 3 - t:
s = 3 - (5/4)
s = 7/4
Now, we can find the intersection point by plugging the value of t or s into one of the original parametric equations. Let's use the first line equation to find the intersection point:
p = (-2, 4, 6) + (5/4)(1, 2, 3)
p = (-2, 4, 6) + (5/4, 5/2, 15/4)
p = (-2 + 5/4, 4 + 5/2, 6 + 15/4)
p = (3/4, 13/2, 39/4)
Therefore, the intersection point of the two lines is (3/4, 13/2, 39/4).
To find the equation of the plane containing the two lines, we can take the cross product of the direction vectors of the lines. Let's call the direction vector of the first line A and the direction vector of the second line B.
A = (1, 2, 3)
B = (3, 2, 1)
The cross product of A and B gives the normal vector of the plane. Let's call it N:
N = A x B
Using the formula for cross product:
N = (2*1 - 3*2, 3*3 - 1*1, 1*2 - 2*3)
N = (-4, 8, -4)
Now, we have the normal vector of the plane. We can use this vector along with any point on the plane to find the equation of the plane. Let's use the intersection point we found earlier, (3/4, 13/2, 39/4), as a point on the plane.
Using the equation of a plane:
Ax + By + Cz = D
Substituting the values:
-4x + 8y - 4z = D
To find D, substitute the coordinates (3/4, 13/2, 39/4) into the equation:
-4(3/4) + 8(13/2) - 4(39/4) = D
-3 + 52 - 39 = D
10 = D
Therefore, the equation of the plane containing the two lines is:
-4x + 8y - 4z = 10.