Determine vector and parametric equations for the plane containing the point P0(1, -2, 3) and having direction vectors a= (4, -2, 5) and b = (-3, 3, 7). Thanks for the help.
How to you define the direction vector of a plane? It it the normal to the plane? How can a plane had two direction vectors?
To determine the vector equation of a plane, we need a point on the plane and the normal vector to the plane.
Given:
Point P0(1, -2, 3)
Direction vectors a = (4, -2, 5) and b = (-3, 3, 7)
To find the normal vector, we can take the cross product of direction vectors a and b.
Normal vector = a × b
= (4, -2, 5) × (-3, 3, 7)
To find the cross product of two vectors, we can use the following formula:
| i j k |
| 4 -2 5 |
= |-3 3 7 |
Using expansion along the first row, we can calculate the cross product as follows:
= ( (-2 * 7) - (5 * 3), (4 * 7) - (5 * -3), (4 * 3) - (-2 * -3) )
= (-14 - 15, 28 + 15, 12 + 6)
= (-29, 43, 18)
So, the normal vector to the plane is N = (-29, 43, 18).
The vector equation of the plane is given by:
N · (r - r0) = 0
where N is the normal vector, r is a general point on the plane, and r0 is a point on the plane.
We can substitute the values into the equation:
(-29, 43, 18) · (r - (1, -2, 3)) = 0
Simplifying this equation, we get:
-29(x - 1) + 43(y + 2) + 18(z - 3) = 0
Expanding further:
-29x + 29 + 43y + 86 + 18z - 54 = 0
-29x + 43y + 18z + 61 = 0
Therefore, the vector equation of the plane is -29x + 43y + 18z + 61 = 0.
To find the parametric equations of the plane, we can set up a system of equations using the vector equation of the plane.
-29x + 43y + 18z + 61 = 0
Let's isolate one variable in terms of others, such that:
-29x = -43y - 18z - 61
Dividing by -29:
x = (43/29)y + (18/29)z + (61/29)
So, the parametric equations of the plane are:
x = (43/29)t + (18/29)u + (61/29)
y = t
z = u
where t and u are parameters.