To find the intersection point of the lines, we can set the two equations equal to each other:
(-2, 4, 6) + t(1, 2, 3) = (-2, 1, 6) + s(3, 2, 1)
This gives us the system of equations:
-2 + t = -2 + 3s
4 + 2t = 1 + 2s
6 + 3t = 6 + s
Simplifying these equations, we get:
t - 3s = 0
2t - 2s = -3
3t - s = 0
We can solve these equations to find the values of t and s. From the first equation, we have t = 3s. Substituting this into the second and third equations, we get:
2(3s) - 2s = -3
6s - 2s = -3
4s = -3
s = -3/4
Substituting this value of s into t = 3s, we get:
t = 3(-3/4) = -9/4
Now we can substitute these values of t and s back into the equation of one of the lines to find the intersection point. Let's use the first line:
p = (-2, 4, 6) + t(1, 2, 3)
p = (-2, 4, 6) + (-9/4)(1, 2, 3)
p = (-2, 4, 6) + (-9/4, -9/2, -27/4)
p = (-2 + (-9/4), 4 + (-9/2), 6 + (-27/4))
p = (-17/4, -1/2, -9/4)
Therefore, the intersection point of the two lines is (-17/4, -1/2, -9/4).
To find the equation of the plane containing the two lines, we can take the cross product of the direction vectors of the two lines. The direction vectors are (1, 2, 3) and (3, 2, 1). Taking the cross product, we get:
(1, 2, 3) x (3, 2, 1) = (4, -8, 4)
This gives us the normal vector of the plane. Using the point (-2, 4, 6) (which lies on the first line), we can write the equation of the plane as:
4(x - (-2)) - 8(y - 4) + 4(z - 6) = 0
4x - 8 + 16y - 32 + 4z - 24 = 0
4x + 16y + 4z - 64 = 0
Therefore, the equation of the plane containing the two lines is 4x + 16y + 4z - 64 = 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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