Hi I wanted to check my work for this problem.
Consider the points A(1,0,0) B(2,2,2) and C(0,2,1).
1) Find the vector CA x CB
I got -2i -3j +4k for this
2) Find an exact value for the area of the triangle ABC.
I got sqrt29/2 for this.
3) Plane II(1) is 2x+3y-4z=2. A second plane II(2) is defined by the Cartesian equation II(2): 4x-y-z=4. Find the vector equation of L1: the intersection of the planes II(1) and II(2).
For this I got L1 = (0, 6, 8) + t(-1,-2,-2) by crossing the two planes, and then setting x to be 0? But I believe that this method might be wrong.
4) A third plane II(3) is defined by the Cartesian equation 16x + ay - 3z = b. Find the value of a if all three planes contain L1.
I got a = 4.
5) Find conditions on a and b if the plane II(3) does not intersect with L1.
I do not know how to tackle this problem at all.
1) To find the cross product between two vectors, CA and CB, you can use the following formula:
CA x CB = (CAy * CBz - CAz * CBy)i - (CAx * CBz - CAz * CBx)j + (CAx * CBy - CAy * CBx)k
Substituting the values:
CA = A - C = (1-0)i + (0-2)j + (0-1)k = i - 2j - k,
CB = B - C = (2-0)i + (2-2)j + (2-1)k = 2i + k
Now, plug these values into the formula:
CA x CB = (2 * (-1) - (-1) * (-2))i - (1 * (-1) - (-1) * 2)j + (1 * 2 - 2 * 2)k
= -2i - 3j + 4k
Therefore, you got the correct answer for the vector CA x CB.
2) To find the area of the triangle ABC, you can use the magnitude of the cross product of vectors CA and CB divided by 2:
Area = |CA x CB| / 2
Substituting the values:
|CA x CB| = |-2i - 3j + 4k| = sqrt((-2)^2 + (-3)^2 + 4^2) = sqrt(4 + 9 + 16) = sqrt(29)
Area = sqrt(29) / 2
So, you got the correct answer for the area of the triangle ABC.
3) To find the vector equation of the intersection line L1 between planes II(1) and II(2), we need to find a point on the line and its direction vector.
First, find a point on the line by solving the system of equations formed by the two planes:
2x + 3y - 4z = 2 (II(1))
4x - y - z = 4 (II(2))
You can solve this system using any method you prefer, such as substitution or elimination. Let's solve it using elimination:
Multiply II(2) by 2:
8x - 2y - 2z = 8 (II(3))
Subtract II(3) from II(1):
-6x + 5y - 2z = -6
Now, we have a system of two equations:
-6x + 5y - 2z = -6
-6x + 9y - 12z = -10
Subtract the first equation from the second equation:
4y - 10z = -4
Now, solve for y in terms of z:
4y = 10z - 4
y = (10z - 4) / 4
y = (5/2)z - 1
Solve for x in terms of z using II(1):
2x + 3(5/2)z - 3 = 2
2x + (15/2)z - 3 = 2
2x + (15/2)z = 5
2x = - (15/2)z + 5
x = (-15/4)z + (5/2)
Now, we have the point (-15/4, 5/2, 0) on the line L1.
Finding the direction vector:
The direction vector of L1 can be found by taking the cross product of the normal vectors of the planes II(1) and II(2).
The normal vector of II(1) is (2, 3, -4).
The normal vector of II(2) is (4, -1, -1).
Taking the cross product of these two vectors gives us the direction vector of L1:
Direction vector = (2i + 3j - 4k) x (4i - j - k) = 3i + 6j + 6k
Therefore, the vector equation of L1 is:
L1 = (-15/4)i + (5/2)j + 0k + t(3i + 6j + 6k)
You made a mistake in the calculation by setting x to 0. Instead, you should follow the steps above to find the correct vector equation of L1.
4) To find the value of a such that all three planes contain L1, you need to substitute the equation of L1 into the Cartesian equation of plane II(3):
16x + ay - 3z = b
By substituting (-15/4)i + (5/2)j + 0k + t(3i + 6j + 6k) for x, y, and z, we get:
16(-15/4) + a(5/2) + 0 = b
Simplifying the equation:
-60 + (5/2)a = b
Now, to satisfy the condition that all three planes contain L1, b must be equal to 0. Thus, we have:
-60 + (5/2)a = 0
Solving this equation for a:
(5/2)a = 60
a = (2/5) * 60
a = 24
Therefore, the value of a is 24.
5) If the plane II(3) does not intersect with L1, it means that L1 is parallel to II(3). For two lines to be parallel, their direction vectors must be scalar multiples of each other.
The direction vector of L1 is (3, 6, 6), which means the direction ratios are 3:6:6.
The direction vector of II(3) is (16, a, -3), which means the direction ratios are 16:a:-3.
For L1 and II(3) to be parallel, the direction ratios of L1 must be proportional to the direction ratios of II(3).
So, we can set up the following proportion:
3:6:6 = 16:a:-3
This gives us the condition:
a/6 = -3/6
a = -3
Therefore, if the plane II(3) does not intersect with L1, then a must be equal to -3.