Find an equation of the plane. The plane through the points (4, 1, 4), (5, -8, 6), and (-4, -5, 1)

tinfoQ

tinfoQ

Answered question

2021-06-04

Find an equation of the plane.
The plane through the points (4, 1, 4), (5, -8, 6), and (-4, -5, 1)

Answer & Explanation

Fatema Sutton

Fatema Sutton

Skilled2021-06-05Added 88 answers

Here it is

Eliza Beth13

Eliza Beth13

Skilled2023-06-19Added 130 answers

Step 1: Find two vectors on the plane.
We can choose P1P2 and P1P3 as two vectors on the plane. The vector P1P2 is given by:
P1P2=54,81,64=1,9,2.
Similarly, the vector P1P3 is given by:
P1P3=44,51,14=8,6,3.
Step 2: Find the normal vector to the plane.
To find the normal vector to the plane, we take the cross product of P1P2 and P1P3:
n=P1P2×P1P3=|𝐢𝐣𝐤192863|
Expanding the determinant, we have:
n=(9)(3)2(6),(1)(3)2(8),(1)(6)(9)(8)=3,3,66.
Step 3: Write the equation of the plane.
Now that we have the normal vector n=3,3,66, we can write the equation of the plane using the point-normal form:
3(xx1)3(yy1)66(zz1)=0, where (x1,y1,z1) is any point on the plane, in this case, we can use P1(4,1,4):
3(x4)3(y1)66(z4)=0.
Simplifying the equation, we get:
3x123y+366z+264=0,
which can be further simplified to:
3x3y66z+255=0.
Therefore, the equation of the plane passing through the points (4,1,4), (5,8,6), and (4,5,1) is 3x3y66z+255=0.
Nick Camelot

Nick Camelot

Skilled2023-06-19Added 164 answers

1. Calculate two vectors using the given points:
v1=54,81,64=1,9,2
v2=44,51,14=8,6,3
2. Find the cross product of v1 and v2 to get the normal vector n of the plane:
n=v1×v2=1,9,2×8,6,3
3. Substitute one of the given points (e.g., (4,1,4)) and the normal vector n into the equation of a plane: Ax+By+Cz=D:
A(4)+B(1)+C(4)=D
4. Solve for D using the coefficients A, B, and C obtained from the normal vector.
The equation of the plane is Ax+By+Cz=D, where the values of A, B, C, and D can be found through the above steps.
Mr Solver

Mr Solver

Skilled2023-06-19Added 147 answers

Result:
(4,1,4), (5,8,6), and (4,5,1)
Solution:
Let's consider vector u defined by the displacement between the first two points: (4,1,4) and (5,8,6). We can express u as follows:
u=54,81,64=1,9,2.
Similarly, vector v is defined by the displacement between the first and third points: (4,1,4) and (4,5,1). We can express v as follows:
v=44,51,14=8,6,3.
Now, we can calculate the cross product of u and v to obtain a vector normal to the plane:
n=u×v=|i^j^k^192863|.
Expanding the determinant, we have:
n=(9)(3)2(6),(2(3)1(8)),(1(6)(9)(8))=15,2,66.
Since the vector n is normal to the plane, we can use it as the coefficients of x, y, and z in the equation of the plane. Let's denote the equation of the plane as ax+by+cz=d, where a, b, c, and d are constants. Substituting the coordinates of any point on the plane, such as (4,1,4), into the equation, we can determine the value of d:
15(4)+2(1)+(66)(4)=dd=60+2264=202.
Therefore, the equation of the plane is:
15x+2y66z=202.
This equation describes the plane passing through the points (4,1,4), (5,8,6), and (4,5,1) in three-dimensional space.

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