Let P=(0,0,1),Q=(1,−1,1),R=(−2,1,−1) . Find (a) The area of the triangle PQR, (b) The equation for a plane that contains P, Q, and R.

Question

Let P=(0,0,1),Q=(1,−1,1),R=(−2,1,−1) . Find

(a) The area of the triangle PQR,

(b) The equation for a plane that contains P, Q, and R.

Answer 1

The first part is easily solved using the Herron Formula : A = sqrt [ s (s -a) ( s-b) (s -c) ]

where s is the semi-perimeter s = .5* (a + b + c) and a , b, and c are the lengths of the sides.

It is straightforward to get a = 1, b = sqrt(3) and c = sqrt( 6)

Plugging in, one gets A = sqrt(1/2)

Part b is more interesting. However more advanced books on linear algebra show that the equation of the plane is given formally by : A dot P = det M

where M is the 3 by 3 matrix

M = Px Py Pz

Qx Qy Qz

Rx Ry Rz

and P is the vector (x,y,z) , and A is the vector P X Q + Q X R + R X P ( here X denotes vector cross product)

dot indicates dot produce and det stands for determinant.

Px = 0, Py = 0, Pz =1 ; Qx = 1 Qy = -1, Qz = 1 ; Rx = -2 Ry = 1 Rz = -1

Notice that det M is just a number and that A dot P will be e x + f y + g z , where e ,f and g are numbers that are obtained from working out the cross products.

So the equation of the plane is e x + g y + g z = det M

I did not have time to work out det M and e,f,g - but it is straightforward.

Answer 2

APPOACH ONE BY USING MEANS OF ANALYTIC GEOMETRY

FOR THE AREA

Let us find the lengths of the triangle.

a =PQ = √2

b = PR = 3 and

c =QR = √17

The semiperimeter of the triangle is S =( 3 + √2 + √17 ) / 2 Then applying Heron's formula

we have A = [ s(s-a )(s-b)(s-c)]^1/2 =[ ( 3 + √2 + √17 ) / 2{ ( 3 + √2 + √17 ) / 2 -√2 }{ ( 3 + √2 + √17 ) / 2-3}{( 3 + √2 + √17 ) / 2-√17}]^1/2

= [( 3 + √2 + √17 ) / 2 ( 3 -√2 + √17 ) / 2 ( -3 + √2 + √17 ) / 2 ( 3 + √2 - √17 ) / 2]^1/2

={ [(3+√17)² - 2][2- (3-√17)²] }^1/2 / 4 = [ (24 +6√17)(-24 +6√17)]^1/2 /4 = 3/2

EQUATION OF THE PLANE

Vector PQ = < 1, -1 ,0 > and vector PR = < -2, 1, -2 >

and their cross product n = < 1, -1 ,0 > x < -2, 1, -2 > = < 2, 2, -1>

Then the equation of the plane is 2 x + 2y - ( z -1) = 0

2 x + 2 y - z = -1

APPOACH TWO BY USING CALCULUS

The surface is given by z = 1 +2 x + 2 y and its projection on the xy- plane

is the region Ω bounded by the lines

y = -x

y = (-1/2) x

y = ( -2/3)x - 1/3

Then the area of the triangle is as follows

∫∫_Ω[ 1 + ( ∂z/∂x)² +( ∂z/∂y)² ]^1/2dA = ∫∫_Ω[ 1 + ( 2)² +( 2)² ]^1/2dA =∫∫_Ω[ 9 ]^1/2dA ∫∫_Ω 3 dA =

3∫₀¹∫_-2x/3-1/3^y=-x dy dx +3 ∫_-2⁰∫_-2x/3-1/3^y=-x/2 dy dx = 1/2 +1 = 3/2