Find the scalar equation of a plane that is perpendicular to plane r=(-2,1,3)+s(0,1,1)+t(-1,0,3) and also intersects it along the line (x-3)/2=(y+1)/-3,z=1
A normal to r is
(0,1,1) cross (-1,0 3)
or
(3, -1, 1) , (I assume you know how to find a cross-product)
So the plane must look like,
3x - y + z = c , where c is a constant
from the line (x-3)/2 = (y+1)/-3 , z=1
the point (3,-1 ,1) must lie on the line, so it must also lie in the plane
3(3) - (-1) + 1 = c
c = 11
equation of your plane is
3x - y + z = 11
To find the scalar equation of a plane perpendicular to another plane and intersecting it along a given line, we need to follow these steps:
Step 1: Find the direction vector of the given line.
Step 2: Find two non-parallel vectors in the given plane to form the normal vector of the desired plane.
Step 3: Use the normal vector and a point on the given line to obtain the scalar equation of the desired plane.
Let's start with Step 1:
Step 1: Finding the direction vector of the line
The given line is expressed as:
(x - 3)/2 = (y + 1)/(-3) = z
Comparing the equation with the standard parametric equation of a line, we can extract the direction vector of the line, which is (2, -3, 1).
Next, let's move on to Step 2:
Step 2: Finding two non-parallel vectors in the given plane
The equation of the given plane is:
r = (-2, 1, 3) + s(0, 1, 1) + t(-1, 0, 3)
By ignoring the parameters "s" and "t" and treating the coefficients as vectors, we can identify two non-parallel vectors in the given plane: u = (0, 1, 1) and v = (-1, 0, 3).
Note that it's important to choose non-parallel vectors to ensure that the normal vector of the desired plane doesn't coincide with the direction vector of the given line.
Now that we have these vectors, let's move on to Step 3:
Step 3: Obtaining the scalar equation of the desired plane
To find the normal vector of the desired plane, we take the cross product of the two vectors from Step 2. Let's denote the normal vector as n:
n = u x v
Calculating the cross product, we have:
n = (0, 1, 1) x (-1, 0, 3) = (3, -3, 1)
Now we have the normal vector of the desired plane, and we need to find a point on the given line. Based on the line equation, we can choose any point that satisfies it. Let's choose the point A(3, -1, 1).
Using the point A(3, -1, 1) and the normal vector n(3, -3, 1), we can write the scalar equation of the desired plane as follows:
n · (r - A) = 0
Substituting the values, we get:
(3, -3, 1) · (x - 3, y + 1, z - 1) = 0
Expanding the dot product, we have:
3(x - 3) - 3(y + 1) + (z - 1) = 0
Simplifying further, we get:
3x - 9 - 3y - 3 + z - 1 = 0
Finally, rearranging the terms, we obtain the scalar equation of the desired plane:
3x - 3y + z - 13 = 0
Therefore, the scalar equation of the plane perpendicular to the given plane and intersecting it along the line (x-3)/2=(y+1)/-3, z=1 is 3x - 3y + z - 13 = 0.