8. Find the vector equation of the line in which the 2 planes 2x - 5y + 3z = 12 and 3x + 4y - 3z = 6 meet.
just take the cross-product of the normals to find the direction of the vector:
| i j k |
| 2 -5 3 |
| 3 4 -3 |
= 3i+15j+23k
Now find a point on the line:
if x=0, y=-18,z=-26
the line is thus
3ti + (-18+15t)j + (-26+23t)k
thanks steve.
do you know how to do these ones? :(
4. Write a vector equation of the line through the point (5, -2, 3) and parallel to the vector v=[4, -3, 1]
5. Determine an equation for the plane that is exactly between the points A(-1, 2, 4) and B(3, 1, -4).
6. Find out if the line r = (1, 3, 8) + t (-2, 5, 7) is parallel to the plane 3x + 4y - 2z = 1
7. Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?
for the #4 i got
r = (5, -2, 3) + t(4, -3, 1)
--- can you please help me on 5, 6, 7. instead since i got 4. thanks steve
To find the vector equation of the line in which the two planes meet, we first need to find the point of intersection between the planes.
To do this, we can set up a system of equations with the two given planes:
2x - 5y + 3z = 12 ...(Equation 1)
3x + 4y - 3z = 6 ...(Equation 2)
We can solve this system of equations by using matrix methods, such as Gaussian elimination or matrix inversion. However, to explain the process step-by-step, we will use the method of substitution.
First, let's isolate one variable in one of the equations. We can choose to isolate "x" in Equation 1. Rearranging Equation 1, we have:
2x = 5y - 3z + 12
x = (5y - 3z + 12)/2 ...(Equation 3)
Now, substitute Equation 3 into Equation 2:
3((5y - 3z + 12)/2) + 4y - 3z = 6
Simplifying this equation, we get:
15y - 9z + 36 + 4y - 3z = 6
19y - 12z + 36 = 6
19y - 12z = -30 ...(Equation 4)
Now, let's isolate another variable. We can isolate "y" in Equation 4:
19y = 12z - 30
y = (12z - 30)/19 ...(Equation 5)
Next, substitute Equation 5 into Equation 3:
x = (5((12z - 30)/19) - 3z + 12)/2
Simplifying this equation, we have:
x = (60z - 150 - 57z + 57z + 228)/38
x = (60z + 78)/38
x = (30z + 39)/19 ...(Equation 6)
Now we have expressions for "x", "y", and "z" in terms of a parameter "z". We can now express the point of intersection between the planes in vector form by combining the expressions:
P = (x, y, z) = [(30z + 39)/19, (12z - 30)/19, z]
Now that we have the point of intersection, we can find the direction vector of the line by subtracting another point on the line from P. A convenient choice for this point would be (0, 0, 0). So, the direction vector can be calculated as:
D = P - (0, 0, 0)
D = [(30z + 39)/19, (12z - 30)/19, z] - (0, 0, 0)
D = [(30z + 39)/19, (12z - 30)/19, z]
Finally, the vector equation of the line in which the two planes meet is:
r = P + tD
Where "P" is the point of intersection and "D" is the direction vector of the line. "r" represents any point on the line and "t" is a scalar variable.