Find an equation of the plane with the given characteristics. The plane passes through the points (4, 2, 1) and (4, 1, -6) and is perpendicular to the plane 8x + 9y+ 3z = 14.
my answer was 57x-56y+8z-124=0 but it was wrong
AAAaannndd the bot gets it wrong yet again!
surely it is obvious that
8x + 9y + 3z = 47
is not perpendicular to
8x + 9y+ 3z = 14
To find the equation of the plane with the given characteristics, we need to consider the normal vector of the plane.
First, let's find the direction vector of the given plane, 8x + 9y + 3z = 14. We can rewrite this equation in the form of Ax + By + Cz = D, where A = 8, B = 9, C = 3, and D = 14.
The normal vector of the plane is then (A, B, C) = (8, 9, 3).
Next, we need to find the direction vector of the new plane. This can be done by taking the difference between the two given points on the new plane, (4, 2, 1) and (4, 1, -6). The direction vector, let's say (a, b, c), is then (4-4, 2-1, 1-(-6)) = (0, 1, 7).
Since the new plane is perpendicular to the given plane, the dot product of their normal vectors should be zero. Using the dot product formula, we have
8a + 9b + 3c = 0.
Substituting the values of a, b, and c from the direction vector of the new plane, we have
8(0) + 9(1) + 3(7) = 0,
0 + 9 + 21 = 0,
30 = 0.
Since 30 is not equal to 0, the equation 57x - 56y + 8z - 124 = 0 is incorrect.
To find the correct equation of the plane, we can use the cross product of the normal vectors of the given plane and the new plane. The cross product of (8, 9, 3) and (0, 1, 7) will give us a normal vector of the new plane.
Taking the cross product, we have:
i j k
8 9 3
0 1 7
= (63 - 9) - (0 - 21)i - (7 - 0)j + (0 - 0)k
= 54i + 7j.
Now, we have the normal vector (54, 7) for the new plane. To find the equation of the plane, we can choose either of the given points on the plane, let's say (4, 2, 1) and substitute the values into the equation of a plane:
54(x - 4) + 7(y - 2) + z - 1 = 0.
Simplifying this equation, we get:
54x - 216 + 7y - 14 + z - 1 = 0,
54x + 7y + z - 231 = 0.
Thus, the correct equation of the plane is 54x + 7y + z - 231 = 0.
To find the equation of a plane, we need to determine a direction vector for the plane and a point that lies on the plane.
First, let's find a direction vector for the given plane. Since the desired plane is perpendicular to the plane 8x + 9y + 3z = 14, the normal vector of the given plane will be perpendicular to the desired plane. Therefore, we can use the coefficients of x, y, and z in the equation of the given plane as the direction vector for the desired plane.
So, the direction vector for the desired plane is <8, 9, 3>.
Next, we need to find a point that lies on the plane. We are given that the plane passes through the points (4, 2, 1) and (4, 1, -6). Either point can be used to find the equation of the plane. Let's choose the point (4, 2, 1).
Using the point (4, 2, 1) and the direction vector <8, 9, 3>, we can write an equation for the desired plane in the form Ax + By + Cz = D. Substituting the values, we have:
8(x - 4) + 9(y - 2) + 3(z - 1) = 0
Expanding this equation, we get:
8x - 32 + 9y - 18 + 3z - 3 = 0
8x + 9y + 3z - 53 = 0
Therefore, the correct equation for the desired plane is 8x + 9y + 3z - 53 = 0, not 57x - 56y + 8z - 124 = 0.
To find the equation of a plane, you need to know its normal vector and a point on the plane.
We are given that the plane is perpendicular to the plane 8x + 9y + 3z = 14. So the normal vector of our plane is parallel to the normal vector of the given plane, which is <8,9,3>.
To find a point on the plane, we can use either of the given points. Let's use (4,2,1):
So now we have the normal vector and a point on the plane, and we can use the point-normal form of the equation of a plane:
<8,9,3> dot (x-4,y-2,z-1) = 0
Expanding the dot product:
8(x-4) + 9(y-2) + 3(z-1) = 0
Simplifying:
8x + 9y + 3z = 47
So the equation of the plane is 8x + 9y + 3z = 47.