Calculate the area of the triangle- Vector Calculus

[chwala]

Homework Statement

see attached.

Relevant Equations

Vector Calculus
$\dfrac {1}{2}$$\left\| {v×w}\right\|$

This is the question,

Now to my question, supposing the vectors were not given, can we let $V=\vec {RQ}$ and $W=\vec {RP}$? i tried using this and i was not getting the required area. Thanks...

 

[Charles Link]

That should give the same answer. Try computing it again.

 

[SammyS]

You should get the same result.

 

[valenumr]

Look up cross product, I think?

 

[chwala]

ok, let me do that again...talk later in the day...I will amend my latex too...laters...

 

[valenumr]

Look up cross product, I think?

I feel silly... The latex rendered slowly.

 

[valenumr]

I already made myself feel silly,but the problem is over specified. You are correct to assume any choice of vector. With only two given, the other is defined, and any choice should give the same answer.

 

[chwala]

Yeah yeah...Great guys same values $155, 190, -29$...i will post working later. Bingo!

 

[chwala]

Yeah yeah...Great guys same values $155, 190, -29$...i will post working later. Bingo!

A=$\dfrac{1}{2}\|(8,-5,10)×(7,-8,-15)\|=\dfrac{1}{2}(155,190,-29)=\dfrac{1}{2}\sqrt{{155^2+190^2+(-29)^2}}≈123.46$

 

[Mark44]

A=$\dfrac{1}{2}\|(8,-5,10)×(7,-8,-15)\|=\dfrac{1}{2}(155,190,-29)=\dfrac{1}{2}\sqrt{{155^2+190^2+(-29)^2}}≈123.46$

Nit: The second expression reads as being 1/2 of the vector <155, 190, -29>. What you should have is 1/2 the norm or magnitude of that vector.

 

[chwala]

Nit: The second expression reads as being 1/2 of the vector <155, 190, -29>. What you should have is 1/2 the norm or magnitude of that vector.

@Mark44 ...you have keen eyes mate! ...I will amend that later in the day...cheers

 

[Mark44]

i tried using this and i was not getting the required area.

That should give the same answer. Try computing it again.

If I'm working a problem and my result doesn't agree with the published answer, the first thing I do is to make sure I'm working the same problem. If that checks out, I then check my work for errors. Although it's possible that the published answer is wrong, this doesn't happen all that often.

 

[chwala]

A=$\dfrac{1}{2}\|(8,-5,10)×(7,-8,-15)\|=\dfrac{1}{2}\|(155,190,-29)\|=\dfrac{1}{2}\sqrt{{155^2+190^2+(-29)^2}}≈123.46$

Amended post $9$.