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.
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
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