Determine the area and the height of the triangle

[iremos]

Hi! i need some help on this question.

In the triangle below, the lengths of OA and OB are LA and LB respectively.
(i) Determine an expression for the length h and an expression for the area of the triangle in terms of θ, LA and LB only.
(ii) Hence, determine the area given OA = i + j + k and OB = i − 2j .

there is what i got so far.

i) h = LBSinθ
Area = LALBSinθ/2

ii) LA = 3^(1/2), LB = 5^(1/2)

How do i solve for the area if i don't know the height and θ?

 

[ksdhart2]

Er... but you do know the area and height, at least in terms of the lengths OA and OB. That's what you found in the first part of the question. Although, I'm confused to as how you've jumped from the given statement that the length of OA is i+j+k to the length of OA is sqrt(3). And I have a similar question about how i - 2j = sqrt(5). Were the values of i, j, and k given elsewhere in the problem text (which you didn't include)? Are they, perhaps, meant to the roots of negative one which define Quaternions? Are they meant to be the standard unit vectors? Please be as specific as you can. Thank you.

 

[iremos]

Given the vector of OA and OB shouldn't the length be the magnitude?
Hence OA = (1^2+1^+1^2)^1/2

same for OB.

 

[ksdhart2]

Oh, okay. If OA and OB are meant to be vectors, and thus i, j, and k are the standard unit (or component) vectors, then, yes, the magnitude of OA would be sqrt(3) and the magnitude of OB would be sqrt(5). I asked for clarification because it wasn't immediately clear to me what i, j, and k meant. In any case, you've got enough information to solve the problem. From your workings in part (i):

\(\displaystyle h=\vert \vert \overrightarrow{OA} \vert \vert \cdot sin(\theta)\)

\(\displaystyle A=\dfrac{\vert \vert \overrightarrow{OA} \vert \vert \cdot \vert \vert \overrightarrow{OB} \vert \vert \cdot sin(\theta)}{2}\)

And \(\displaystyle sin(\theta)\) can be calculated via the magnitude of the cross product:

\(\displaystyle \vert \vert \overrightarrow{OA} \times \overrightarrow{OB} \vert \vert = \vert \vert \overrightarrow{OA} \vert \vert \cdot \vert \vert \overrightarrow{OB} \vert \vert \cdot sin(\theta)\)