Find an equation of the plane that passes through the point (1, 2,3) and cuts off the smallest volume in the first octant. [This one is a little tricky.]

Answer and Step-by-step explanation:

Solution:

The equation is given by:

z – z0 = a(x – x0) + b(y – y0).

The point (1, 2, 3) is given , substitute them for x0, y0 and z0:

Z – 3 = a (x – 1) + b(y-2)

Both a and b should be negative, because normal vector to point in the negative x and y direction so that plane will enclose a tetrahedron with finite area in the first octant.

This will only happen if z becomes smaller as we move towards the positive x and y directions.

To find expressions for V, firstly find expression for x, y and z intercepts.

X – Intercepts (y and z = 0) : 2b +a -3 / a

y- Intercepts (x, z = 0): 2b + a- 3 / b

z – Intercepts( x, y = 0): 3 – a – 2b

The base is = (2b+a-3)2/ 2ab

Volume is 1/3 base x height, so

V (a, b) = - (2b+a-3)3 / 6ab

The gradient vector:

ΔV (a, b) = |- (3(2b+a-3)2 ab-b (2b+a-3)3)/6a2b2, (6(2b+a-3)2 ab-a(2b+a-3)3)/6a2b2| = 0

b (2b+a-3)2 [ (2b+a-3) -3a] = 0

a (ab + a-3)2 [(2b +a -3) – 6b] = 0

Where a, b not equal to 0.

We cannot solve for 2b +a-3 = 0 because, that will lead to V = 0, which is impossible.

Therefore, we only solve for second halves equations, which are system of linear equations:

-2a + 2b = 3 and a – 4b = 3

Which gives a = -3 and b = -3/2

Put in equation:

Z – 3 = -3 (x-1) – 3/2 (y-2)

3x +3/2 y +z = 9

Multiplying by 2:

6x + 3y + 2z = 18.

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)

black/brown

step-by-step explanation:

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