# Vectors/C4

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Thread starter 4 years ago
#1
Hi!

I have a doubt to clarify regarding the following question(Question 13 and Exercise 5G from the Edexcel Coursebook):-

Given that the points A and B have coordinates (7,4,4) and (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 the triangle is 5 multiplied by the square root of 29 divided by 2.

I was able to find the value of cos AOB and I'm aware on how to find the area of the triangle but when I solved the question, I wasn't able to prove the area as per given in the question, I checked the solution bank and they used the trigonometric identity sin2 (theta)+cos2(theta)= 1 after finding the value of cos AOB. Why did they do so? Is there any reason behind it?

http://pmt.physicsandmathstutor.com/...hapter%205.pdf
0
4 years ago
#2
(Original post by sabahshahed294)
Hi!

I have a doubt to clarify regarding the following question(Question 13 and Exercise 5G from the Edexcel Coursebook):-

Given that the points A and B have coordinates (7,4,4) and (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 the triangle is 5 multiplied by the square root of 29 divided by 2.

I was able to find the value of cos AOB and I'm aware on how to find the area of the triangle but when I solved the question, I wasn't able to prove the area as per given in the question, I checked the solution bank and they used the trigonometric identity sin2 (theta)+cos2(theta)= 1 after finding the value of cos AOB. Why did they do so? Is there any reason behind it?

http://pmt.physicsandmathstutor.com/...hapter%205.pdf
Maybe because of this formula for the area of a triangle, .
0
4 years ago
#3
(Original post by sabahshahed294)
Hi!

I have a doubt to clarify regarding the following question(Question 13 and Exercise 5G from the Edexcel Coursebook):-

Given that the points A and B have coordinates (7,4,4) and (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 the triangle is 5 multiplied by the square root of 29 divided by 2.

I was able to find the value of cos AOB and I'm aware on how to find the area of the triangle but when I solved the question, I wasn't able to prove the area as per given in the question, I checked the solution bank and they used the trigonometric identity sin2 (theta)+cos2(theta)= 1 after finding the value of cos AOB. Why did they do so? Is there any reason behind it?

http://pmt.physicsandmathstutor.com/...hapter%205.pdf
They did it to find the value of and use the formula mentioned above.
0
Thread starter 4 years ago
#4
(Original post by GPiph)
Maybe because of this formula for the area of a triangle, .
(Original post by RDKGames)
They did it to find the value of and use the formula mentioned above.
But then, why can't we just use the value of the angle that we calculated? Why do we exactly have to calculate the value of sin theta?
0
4 years ago
#5
(Original post by sabahshahed294)
But then, why can't we just use the value of the angle that we calculated? Why do we exactly have to calculate the value of sin theta?
If you know a formula for the area of a triangle concerning cosine of the angle, then by all means go for it. If you don't, then you need to find and use sine of the angle.
0
Thread starter 4 years ago
#6
(Original post by RDKGames)
If you know a formula for the area of a triangle concerning cosine of the angle, then by all means go for it. If you don't, then you need to find and use sine of the angle.
Oh okay. Thank you.
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Thread starter 4 years ago
#7
(Original post by RDKGames)
...
May I ask you another question if you don't mind regarding direction vectors? Considering the given case,

Points A,B,C and D in a plane have position vectors a= 6i + 8j , b= 3/2 a , c= 6i + 3j and d= 5/3 c respectively. Write down the vector equations of the lines AD and BC and find the position vector of their point of intersection.

Now, I had found the vector positions of AD and BC but I was wondering that can the direction vectors change? I could understand in this case the direction vector but the direction vector will change right if let's say we write it as CB and DA?
0
4 years ago
#8
(Original post by sabahshahed294)
May I ask you another question if you don't mind regarding direction vectors? Considering the given case,

Points A,B,C and D in a plane have position vectors a= 6i + 8j , b= 3/2 a , c= 6i + 3j and d= 5/3 c respectively. Write down the vector equations of the lines AD and BC and find the position vector of their point of intersection.

Now, I had found the vector positions of AD and BC but I was wondering that can the direction vectors change? I could understand in this case the direction vector but the direction vector will change right if let's say we write it as CB and DA?
The direction vector of CB is -1 multiplied by the direction vector of BC as you're going from C to B this time as opposed to from B to C hence in the opposite direction - so yes the direction vector is different. However the vector equation of the line would be correct in both cases as long as your fixed point is a point on the line, so as a basic example.
0
Thread starter 4 years ago
#9
(Original post by RDKGames)
The direction vector of CB is -1 multiplied by the direction vector of BC as you're going from C to B this time as opposed to from B to C hence in the opposite direction - so yes the direction vector is different. However the vector equation of the line would be correct in both cases as long as your fixed point is a point on the line, so as a basic example.
Alright. Thank you for your time!
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