Vector - Prove that the area of ∆AOB is 5√29/2.?

Question

Given that the points A and B have coordinates (7, 4, 4) & (2, -2, -1) respectively, use a vector method to find the value of cos AOB, where O is the origin.
Prove that the area of ∆AOB is 5√29/2.

I’ve done 1st portion of this problem. But in second portion, how can I calculate the area of ∆AOB?

Answer 1

The area of a triangle can be calculated using the length of two sides and the included angle.

Area = (1/2)ab sinθ
where θ is the included angle AOB

Calculate the cross product of vectors A and B.
A X B = <7, 4, 4> X <2, -2, -1> = <4, 15, -22>

Calculate the magnitude of the cross product.
|| A X B || = √[4² + 15² + (-22)²] = √(16 + 225 + 484)
|| A X B || = √725 = 5√29

The magnitude of cross product can also be defined as:

|| A X B || = || A || || B || sinθ
where θ is the included angle

Notice that this is just twice the area formula for the triangle.

Area = (1/2) || A || || B || sinθ = (1/2) || A X B ||
Area = (1/2) (5√29) = (5√29) / 2

Answer 2

One way to tackle this is to use "Heron's formula" for calculating the area of a triangle if you know the length of each side. We can determine the length of each side by determining the distance between points.

From O to A = Sqrt(7^2 + 4^2 + 4^2) = 9
From O to B = Sqrt(2^2 + 2^2 + 1^2) = 3
From A to B = Sqrt(5^2 + 6^2 + 5^2) = Sqrt(86)=9.273618495


Now Heron's formula for area = Sqrt(s(s-a)(s-b)(s-c)) where s = 1/2(a+b+c). a, b and c are the length of the sides of the triangle.

So, s=.5(9+3+9.273618495) = 10.63680925
Area = Sqrt(s(s-9)(s-3)(s-9.273618495)
Area = Sqrt(181.25) = 13.4629

This agrees with your answer.

Answer 3

OA * OB = |0A| |OB| cosθ
2 = √81 * √9 cosθ
cosθ = 2/(27)

sinθ = √(725) / (27) = 5√29 / 27

Area = 1/2 |0A| |OB| sinθ
= 1/2 * 9 * 3 * 5 * √29 / (27)
= 5/2 * √29

Answer 4

The cross product of the vectors OA and OB is a vector whose length is equal to the area of the parallelogram defined by OA and OB. Divide that length by 2 and you have the area of the triangle.

Answer 5

Part 1
Let θ = /_AOB
a . b = | a | | b | cos θ
cos θ = (14 - 8 - 4) / (√ 101) (√9 )
cos θ = (2/3) / √101

Part 2
A = (1/2) | a x b |
A = (1/2) |i __j__k|
________|7_4__4|
________|2_-2_-1|

A = 1/2 | 4i + 15j - 22k|
A = 1/2 (4² + 15² + 22²)^(1/2)
A = (1/2)(725)^1/2
A = (5/2) √ 29