Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

Question

delilnaT · Accepted Answer

Let the given points be P(2,1,2),Q(3,-8,6), and R(-2,-3,1) PQ→=⟨3−2,−8−1,6−2⟩=⟨1,−9,4⟩ PR→=⟨−2−2,−3−1,1−2⟩=⟨−4,−4,−1⟩ PQ→×PR→=|ijk1−94−4−4−1|=(9+16)i+(−16+1)j+(−4−36)k =25i−15j−40k Therefore, the normal vector to the plane is n→=⟨25,−15,−40⟩ Since, the plane passes through all the three points we can choose any point to find its equation. So, the equation of the plane through the point P(2,1,2) with normal vector n→=⟨25,−15,−40⟩ is 25(x−2)−15(y−1)−40(z−2)=0 ⇒25x−50−15y+15−40z+80=0 ⇒25x−15y−40z+45=0

alenahelenash · Accepted Answer

To find an equation of the plane passing through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1), we can use the concept of cross products and the equation of a plane.Let&#039;s denote the three given points as P1(2,1,2), P2(3,−8,6), and P3(−2,−3,1). We need to find the normal vector of the plane, which can be obtained by taking the cross product of two vectors lying on the plane.Let&#039;s consider the vectors v1→ and v2→ defined as follows:v1→=P2P1→=[3−2−8−16−2]=[1−94],v2→=P2P3→=[−2−3−3−(−8)1−6]=[−55−5].To find the normal vector n→ of the plane, we take the cross product of v1→ and v2→:n→=v1→×v2→=[1−94]×[−55−5].Using the determinant expansion method, we can calculate the cross product as follows:n→=[(−9)(−5)−4(5)4(−5)−1(−5)1(5)−(−9)(−5)]=[−35−1540].Now, we have the normal vector n→=[−35−1540]. To find the equation of the plane, we can use the point-normal form of the equation, given by:Ax+By+Cz=D,where (A,B,C) is the normal vector and (x,y,z) represents any point on the plane.Let&#039;s substitute the coordinates of one of the given points, such as P1(2,1,2), into the equation. This allows us to determine the value of D:−35(2)−15(1)+40(2)=D,Simplifying, we have:−70−15+80=D,−5=D.Thus, the equation of the plane becomes:−35x−15y+40z=−5.

star233 · Accepted Answer

Step 1. Find the vectors 𝐮 and 𝐯 using two of the given points:𝐮=PQ=[3−2−8−16−2]=[1−94],𝐯=PR=[−2−2−3−11−2]=[−4−4−1].Step 2. Calculate the cross product 𝐧=𝐮×𝐯:𝐧=[1−94]×[−4−4−1]=[31−1540].Step 3. The equation of the plane can be written as Ax+By+Cz=D, where 𝐧=[ABC] is the normal vector to the plane. Substituting one of the given points, such as P(2,1,2), we can solve for D:31(2)−15(1)+40(2)=D⟹62−15+80=D⟹D=127.Step 4. Therefore, the equation of the plane is 31x−15y+40z=127.

karton · Accepted Answer

Result:−52x+15y−40z+169=0Solution:Let&#039;s assume the equation of the plane is Ax+By+Cz+D=0, where A, B, C are the coefficients of the variables x, y, z, respectively, and D is a constant.First, we need to find the normal vector to the plane. We can obtain the normal vector by taking the cross product of two vectors in the plane. We can choose two vectors as follows:v1→=(3−2,−8−1,6−2)=(1,−9,4)v2→=(−2−2,−3−1,1−2)=(−4,−4,−1)To find the normal vector, we calculate the cross product:N→=v1→×v2→=|i→j→k→1−94−4−4−1|Expanding the determinant, we have:N→=(−36−16)i→−(1−16)j→−(4+36)k→=−52i→+15j→−40k→Now, we can use one of the given points (let&#039;s choose (2, 1, 2)) to find the value of D in the equation of the plane. Substituting the coordinates of the point into the equation, we have:−52(2)+15(1)−40(2)+D=0−104+15−80+D=0−169+D=0D=169Therefore, the equation of the plane is:−52x+15y−40z+169=0

Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

Answered question

Answer & Explanation

New Questions in Trigonometry