Write the vector equation and the parametric equations of the plane that is perpendicular to the line [x,y,z] = [5,3,-1]+t[-1,7,2] and parallel to the line [x,y,z] = [2,-3,1]+s[-1,3,-2] and containing the point (1, 2, 5).
To find the vector equation and parametric equations of the plane that is perpendicular to one line and parallel to another line while containing a given point, you can follow these steps:
Step 1: Find the direction vectors of both lines.
For the line [x,y,z] = [5,3,-1]+t[-1,7,2], the direction vector is [-1,7,2].
For the line [x,y,z] = [2,-3,1]+s[-1,3,-2], the direction vector is [-1,3,-2].
Step 2: Find the normal vector of the plane.
To find a vector that is perpendicular to both direction vectors, you can take their cross product.
Let's calculate the cross product of [-1,7,2] and [-1,3,-2]:
[-1, 7, 2] × [-1, 3, -2] = (7 * (-2) - 2 * 3, -1 * (-2) - 2 * (-1), -1 * 3 - 7 * (-1))
= (-8, 0, -10)
So, the normal vector of the plane is [-8, 0, -10].
Step 3: Write the vector equation of the plane.
The vector equation of a plane is given by the formula: n · (r - r0) = 0, where n is the normal vector, r is a position vector on the plane, and r0 is a known point on the plane.
Let's use the point (1, 2, 5) as r0.
So, the vector equation of the plane is: [-8, 0, -10] · ([x, y, z] - [1, 2, 5]) = 0
Simplifying this equation, we get: -8(x-1) - 10(z-5) = 0.
Step 4: Convert the vector equation to parametric equations.
To convert the vector equation to parametric equations, you can solve for one variable, preferably z, in terms of the other variables.
-8(x-1) - 10(z-5) = 0
Expanding and rearranging, we get: -8x + 8 - 10z + 50 = 0
Simplifying, we get: -8x - 10z + 58 = 0
Rearranging, we have: z = -0.8x + 5.8
So, the parametric equations of the plane are:
x = t
y = s
z = -0.8t + 5.8, where t and s are parameters.