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

Question

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

Fatema Sutton · Accepted Answer

Here it is

Eliza Beth13 · Accepted Answer

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→=⟨5−4,−8−1,6−4⟩=⟨1,−9,2⟩.Similarly, the vector P1P3→ is given by:P1P3→=⟨−4−4,−5−1,1−4⟩=⟨−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→=|𝐢𝐣𝐤1−92−8−6−3|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(x−x1)−3(y−y1)−66(z−z1)=0, where (x1,y1,z1) is any point on the plane, in this case, we can use P1(4,1,4):3(x−4)−3(y−1)−66(z−4)=0.Simplifying the equation, we get:3x−12−3y+3−66z+264=0,which can be further simplified to:3x−3y−66z+255=0.Therefore, the equation of the plane passing through the points (4,1,4), (5,−8,6), and (−4,−5,1) is 3x−3y−66z+255=0.

Nick Camelot · Accepted Answer

1. Calculate two vectors using the given points:v1→=⟨5−4,−8−1,6−4⟩=⟨1,−9,2⟩v2→=⟨−4−4,−5−1,1−4⟩=⟨−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)=D4. 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 · Accepted Answer

Result:(4,1,4), (5,−8,6), and (−4,−5,1)Solution:Let&#039;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→=⟨5−4,−8−1,6−4⟩=⟨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→=⟨−4−4,−5−1,1−4⟩=⟨−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^1−92−8−6−3|.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&#039;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)=d⟹d=60+2−264=−202.Therefore, the equation of the plane is:15x+2y−66z=−202.This equation describes the plane passing through the points (4,1,4), (5,−8,6), and (−4,−5,1) in three-dimensional space.

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

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