# Vectors/C4

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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 sin

http://pmt.physicsandmathstutor.com/...hapter%205.pdf

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 sin

^{2 }(theta)+cos^{2}(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

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(Original post by

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 sin

http://pmt.physicsandmathstutor.com/...hapter%205.pdf

**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 sin

^{2 }(theta)+cos^{2}(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

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#3

**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 sin

^{2 }(theta)+cos

^{2}(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

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#5

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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?

**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?

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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.

**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.

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...

**RDKGames**)...

Points A,B,C and D in a plane have position vectors a= 6

**i**+ 8

**j**, b= 3/2 a , c= 6

**i**+ 3

**j**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?

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#8

(Original post by

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= 6

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?

**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= 6

**i**+ 8**j**, b= 3/2 a , c= 6**i**+ 3**j**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?

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(Original post by

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.

**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.

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