Find an equation for the plane consisting of all points that are equidistant from the points (1,0,-2) and (3,4,0) .

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Corben Pittman · Accepted Answer

Start by defining the two endpoints. P=(1,0,−2) Q=(3,4,0) Calculate the midpoint M, which is just the average of the x coordinates, y coordinates, and z coordinates of P and Q. This point is equidistant between P and Q, and runs along the line between them. Since it is equidistant, it will lie in the plane. M=12(1+3,0+4,−2+0)=(2,2,−1) Calculate a normal vector n, in the direction of the line between P and Q. The plane is perpendicular to this line, so this line can function as our normal vector. n=PQ=&amp;lt;2,4,2&amp;gt; Write an equation for the plane based on the normal vector n and the midpoint M. plane: 2x+4y+2z=2(2)+4(2)+2(−1) Simplify. 2x+4y+2z=10 Simplify. x+2y+z=5 Result: x+2y+z=5

user_27qwe · Accepted Answer

Let&#039;s denote a general point on the plane as (x,y,z). We know that this point is equidistant from (1,0,−2) and (3,4,0).Using the distance formula, the distance between a point (x,y,z) and (1,0,−2) is given by:(x−1)2+(y−0)2+(z−(−2))2Similarly, the distance between the same point (x,y,z) and (3,4,0) is given by:(x−3)2+(y−4)2+(z−0)2Since the point (x,y,z) is equidistant from both points, we can equate these two distances:(x−1)2+y2+(z+2)2=(x−3)2+(y−4)2+z2Squaring both sides of the equation to eliminate the square roots, we have:(x−1)2+y2+(z+2)2=(x−3)2+(y−4)2+z2Expanding and simplifying, we get:x2−2x+1+y2+z2+4z+4=x2−6x+9+y2−8y+16+z2Simplifying further, we obtain:−4x−8y−4z+22=0Thus, the equation for the plane consisting of all points equidistant from (1,0,−2) and (3,4,0) is:−4x−8y−4z+22=0

star233 · Accepted Answer

Step 1. Find the midpoint M of the line segment connecting the two points. The midpoint can be calculated by taking the average of the coordinates of the two points.M=(1+32,0+42,−2+02)=(2,2,−1)Step 2. Find the direction vector d→ of the line connecting the two points. The direction vector can be obtained by subtracting the coordinates of one point from the other.d→=⟨3−1,4−0,0−(−2)⟩=⟨2,4,2⟩Step 3. Find a vector v→ that is perpendicular to d→ and can lie in the plane. We can choose any vector that is orthogonal to d→. One possible choice is v→=⟨−4,2,0⟩.Step 4. Use the coordinates of the midpoint M and the vector v→ to write the equation of the plane in vector form.n→·(r→−M→)=0 where n→ is the normal vector of the plane, given by n→=d→×v→ (the cross product of d→ and v→), and r→ represents any point (x,y,z) on the plane.Step 5. Convert the equation to scalar form. Letting r→=⟨x,y,z⟩, we have:n→·(r→−M→)=0⟹⟨2,4,2⟩·(⟨x,y,z⟩−⟨2,2,−1⟩)=0Expanding this equation gives:2(x−2)+4(y−2)+2(z+1)=0Simplifying further:2x+4y+2z−10=0Therefore, the equation of the plane consisting of all points equidistant from (1,0,−2) and (3,4,0) is 2x+4y+2z−10=0.

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