Wrtie a scalar equation given it's parametric equation.
x = 3-2s+2t
y = 1+3s+t
z = 5-s-2t
Two direction vectors on the plane are (-2,3,-1) and (2,1,-2)
A normal to these is (5,6,8), their cross-product
so the plane has equation
5x + 6y + 8z = C
but (3,1,5) is a point on it, so
15 + 6 + 40 = C = 61
equation: 5x + 6y + 8z = 61
To write a scalar equation given the parametric equations, we need to eliminate the parameters (s and t) and express x, y, and z in terms of a single variable.
Let's start by isolating variables s and t in each equation:
1. From x = 3 - 2s + 2t, we can rearrange to get s = (3 - x + 2t) / 2
2. From y = 1 + 3s + t, we can rearrange to get s = (y - 1 - t) / 3
3. From z = 5 - s - 2t, we can rearrange to get s = (5 - z + 2t)
Since they all represent s, we can equate the expressions:
(3 - x + 2t) / 2 = (y - 1 - t) / 3 = (5 - z + 2t)
To find a common expression for x, y, and z, we can eliminate t by setting the two pairs equal to each other:
(3 - x + 2t) / 2 = (5 - z + 2t)
Cross multiplying gives:
2(3 - x + 2t) = 3(5 - z + 2t)
Simplifying the left side gives:
6 - 2x + 4t = 15 - 3z + 6t
Simplifying the right side gives:
6 - 2x + 4t = 15 - 3z + 6t
Combining like terms gives:
-2x + 4t = -3z + 9
This equation: -2x + 4t = -3z + 9, represents the scalar equation for the given parametric equations x = 3 - 2s + 2t, y = 1 + 3s + t, and z = 5 - s - 2t.
To write a scalar equation given its parametric equations, we need to eliminate the parameters (s and t) and express x, y, and z in terms of a single variable, let's say "k".
We have the following parametric equations:
x = 3 - 2s + 2t ...(1)
y = 1 + 3s + t ...(2)
z = 5 - s - 2t ...(3)
To eliminate s and t, let's solve equations (1) and (3) for s and t in terms of x, y, and z:
From equation (1), we have:
2s - 2t = 3 - x
s - t = (3 - x)/2 ...(4)
From equation (3), we have:
s + 2t = 5 - z
s = 5 - z - 2t ...(5)
Substituting the value of s from equation (5) into equation (4), we can solve for t:
(5 - z - 2t) - t = (3 - x)/2
5 - z - 3t = (3 - x)/2
6 - 2z - 6t = 3 - x
6t = 2z - x - 3
t = (2z - x - 3)/6 ...(6)
Now, substituting the solved value of t from equation (6) into equation (5), we can solve for s:
s = 5 - z - 2((2z - x - 3)/6)
s = 5 - z - (4z - 2x - 6)/6
s = 5 - z - 4z/6 + x/3 - 1
s = 4 - 11z/6 + x/3 ...(7)
Finally, we can substitute the values of s and t from equations (6) and (7) into equation (2) to get y in terms of x, y, and z:
y = 1 + 3(4 - 11z/6 + x/3) + (2z - x - 3)/6
y = 1 + 12 - 11z/2 + x/1 - 9/2 + z/3 - x/6 - 1/2 + 3z/6
y = 4 - 8z/6 + 5x/6 ...(8)
Now, combining equations (6), (7), and (8), we have a scalar equation involving x, y, and z:
x + 3y + 2z = x + 3(4 - 8z/6 + 5x/6) + 2z
x + 3y + 2z = x + 12 - 8z/2 + 5x/2 + 2z
x + 3y + 2z = 12 + 4x - 4z + 4z
x + 3y + 2z = 12 + 4x
Therefore, the scalar equation is:
x + 3y + 2z = 12 + 4x