Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0. I know that normal of the latter equation is (3,2,6) but now what do I do? Thanks.
bro its 2018 dab on them haters
I appreciate the help, but you made another mistake. You got a wrong answer, it should be (2,0,-4) not (2,0,-1) so thus, you used the wrong information. But thanks, I know what to do now :)
my line
<but M(1,2,3) lies on it, so 2 - 12 - 1 = k >
should say :
but M(1,2,3) lies on it, so 2 - 12 + 3 = k
Well, it seems like you're stuck between planes! Don't worry, I'll help you out with a funny solution.
To find the scalar equation of the plane through M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z + 1 = 0, you're on the right track by finding the normal vector.
Since the plane you're looking for is perpendicular to the given plane, the normal vector of the desired plane will be parallel to the normal vector of the given plane, which is indeed (3,2,6).
Now, let's use the point normal form of a plane equation to get our answer.
The equation of a plane is given by Ax + By + Cz + D = 0, where (A,B,C) is the normal vector of the plane.
Using the point-normal form equation, we can substitute the coordinates of one of the points, let's say M(1,2,3), into the equation to solve for D.
Substituting M into the equation, we have:
3(1) + 2(2) + 6(3) + D = 0
Simplifying this equation, we get:
3 + 4 + 18 + D = 0
25 + D = 0
D = -25
Therefore, the scalar equation of the plane you're looking for is:
3x + 2y + 6z - 25 = 0
There you have it! A clown-friendly solution to finding the scalar equation of the plane. Hope it made you smile!
To find the scalar equation of the plane passing through points M(1,2,3) and N(3,2,-1) and perpendicular to the plane with equation 3x + 2y + 6z + 1 = 0, you can follow these steps:
Step 1: Find the direction vector of the line passing through points M and N. This can be done by subtracting the coordinates of the points:
Direction vector = N - M = (3-1, 2-2, -1-3) = (2, 0, -4)
Step 2: Find the cross product of the direction vector and the normal vector of the given plane. Since the plane is perpendicular to the given plane, the cross product should be perpendicular to both the direction vector and the normal vector.
Given normal vector = (3, 2, 6)
Cross product = (2, 0, -4) x (3, 2, 6) = (-8, -4, 4)
Step 3: Use the coordinates of one of the given points, let's say M(1,2,3), and the values of the cross product to find the equation of the plane. The equation of the plane can be expressed as:
Ax + By + Cz + D = 0
Substitute the coordinates and the cross product:
-8(x - 1) - 4(y - 2) + 4(z - 3) = 0
Simplify the equation:
-8x + 8 - 4y + 8 + 4z - 12 = 0
-8x - 4y + 4z - 4 = 0
Divide all terms by -4 to write the scalar equation:
2x + y - z + 1 = 0
So, the scalar equation of the plane passing through points M(1,2,3) and N(3,2,-1) and perpendicular to the plane with equation 3x + 2y + 6z +1 = 0 is 2x + y - z + 1 = 0.
You are on the right track, so (3,2,6) must be a direction vector on your new plane.
you also have two points M(1,2,3) and N(3,2,-1), so the vector MN or (2,0,-1) is another direction vector.
So by taking the cross-product of vectors (3,2,6) and (2,0,-1) I got (2,-6,1)
which must be the normal to the new plane.
So the new plane has equation
2x - 6y + z = k
but M(1,2,3) lies on it, so 2 - 12 - 1 = k
k = -7
The new plane has equation 2x - 6y + x = -7
Check:
1. Both M and N satisfy the new equation
2. Is the dot-product of their normals zero? (3,2,6)∙(2,-6,1) = 6-12+6 = 0
There you go!