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 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.
To find the intersection point of the two lines, we need to equate their equations and solve for the parameter values at which they intersect.
The equations of the lines are:
Line 1: p = (-2, 4, 6) + t(1, 2, 3)
Line 2: p = (-2, 1, 6) + t(3, 2, 1)
Setting the coordinates equal to each other:
-2 + t1 = -2 + 3t2 --> t1 = 3t2
4 + 2t1 = 1 + 2t2 --> 4 + 2(3t2) = 1 + 2t2
4 + 6t2 = 1 + 2t2 --> 4t2 = -3
t2 = -3/4
Substituting t2 back into t1 = 3t2:
t1 = 3(-3/4) = -9/4
Now we can find the intersection point by plugging the values of t1 and t2 into one of the line equations. Let's use Line 1:
p = (-2, 4, 6) + t1(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, -3/4)
So, the intersection point of the two lines is (-17/4, -1/2, -3/4).
To find the equation of the plane containing the two lines, we need a vector that is parallel to both lines. This can be found by taking the cross product of the direction vectors of the two lines.
Direction vector of Line 1: (1, 2, 3)
Direction vector of Line 2: (3, 2, 1)
Taking the cross product:
Cross product = (2*1 - 3*2, 3*3 - 1*1, 1*2 - 3*2)
= (-4, 8, -4)
Now we have a vector (-4, 8, -4) that is parallel to the plane. We can use this vector and the coordinates of one of the lines to write the equation of the plane.
Using Line 1:
Point on the plane = (-2, 4, 6)
Equation of the plane:
-4(x - (-2)) + 8(y - 4) - 4(z - 6) = 0
-4(x + 2) + 8(y - 4) - 4(z - 6) = 0
-4x - 8 + 8y - 32 - 4z + 24 = 0
-4x + 8y - 4z - 16 = 0
So, the equation of the plane containing the two lines is -4x + 8y - 4z - 16 = 0.