parallelogram PQRS.

Given that PQ=i+10j-3k and QR=8i+3j+2k, find the area of the quadrilateral, to one decimal place. (6 marks) (from Edexcel practice book)

I am getting 92.0, but the answer is 86.3. I just did the modulus of both then times them together? Seems like that way isn't correct.

Does anyone know how to tackle this problem?

Given that PQ=i+10j-3k and QR=8i+3j+2k, find the area of the quadrilateral, to one decimal place. (6 marks) (from Edexcel practice book)

I am getting 92.0, but the answer is 86.3. I just did the modulus of both then times them together? Seems like that way isn't correct.

Does anyone know how to tackle this problem?

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Original post by sebi12345

parallelogram PQRS.

Given that PQ=i+10j-3k and QR=8i+3j+2k, find the area of the quadrilateral, to one decimal place. (6 marks) (from Edexcel practice book)

I am getting 92.0, but the answer is 86.3. I just did the modulus of both then times them together? Seems like that way isn't correct.

Does anyone know how to tackle this problem?

Given that PQ=i+10j-3k and QR=8i+3j+2k, find the area of the quadrilateral, to one decimal place. (6 marks) (from Edexcel practice book)

I am getting 92.0, but the answer is 86.3. I just did the modulus of both then times them together? Seems like that way isn't correct.

Does anyone know how to tackle this problem?

If you multiplied, that would assume it's a rectangle?

How to adapt for a parallelogram?

Is the book for the current a-level (further)?

(edited 4 years ago)

Original post by mqb2766

If you multiplied, that would assume it's a rectangle?

How to adapt for a parallelogram?

Is the book for the current a-level (further)?

How to adapt for a parallelogram?

Is the book for the current a-level (further)?

yes current a level maths

image is attached QUESTION 6

Original post by sebi12345

yes current a level maths

image is attached QUESTION 6

image is attached QUESTION 6

Ok so what is the area of a parallelogram?

Original post by mqb2766

Ok so what is the area of a parallelogram?

base times vertical height

Original post by sebi12345

Oh base times VERTICAL height. ok let me try it again.

Yes, you'll need the angle between the vectors to get the vertical height.

Have you done the dot product (further)?

Original post by mqb2766

Yes, you'll need the angle between the vectors to get the vertical height.

Have you done the dot product (further)?

Have you done the dot product (further)?

yes, doing year 1 further. Ok ill try that. Shouldnt there be another way as this question isnt further?

Original post by sebi12345

yes, doing year 1 further. Ok ill try that. Shouldnt there be another way as this question isnt further?

got 68.6 degrees. Then what?

Original post by sebi12345

yes, doing year 1 further. Ok ill try that. Shouldnt there be another way as this question isnt further?

You can split into two triangles and for each triangle you can find all three lengths, then find an angle and use the 1/2 ab sin(C) area formula.

Original post by sebi12345

yes, doing year 1 further. Ok ill try that. Shouldnt there be another way as this question isnt further?

The dot is the easiest. It used to be normal maths, hence the earlier question.

Original post by sebi12345

got 68.6 degrees. Then what?

|a||b|sin(theta)

(edited 4 years ago)

I am not sure but about normal maths but if you do FP1 there is very very easy method do you want me to share it?

Original post by sebi12345

parallelogram PQRS.

Given that PQ=i+10j-3k and QR=8i+3j+2k, find the area of the quadrilateral, to one decimal place. (6 marks) (from Edexcel practice book)

I am getting 92.0, but the answer is 86.3. I just did the modulus of both then times them together? Seems like that way isn't correct.

Does anyone know how to tackle this problem?

Given that PQ=i+10j-3k and QR=8i+3j+2k, find the area of the quadrilateral, to one decimal place. (6 marks) (from Edexcel practice book)

I am getting 92.0, but the answer is 86.3. I just did the modulus of both then times them together? Seems like that way isn't correct.

Does anyone know how to tackle this problem?

I may be wrong but I wouldn't expect to see a question like this in the final exam because the further maths dot product method is the quicker approach (cross product is even faster if you do further pure). A Level maths questions are normally designed so that on spec methods are the quickest and most efficient.

i am bored so here it is...

screen shot at 48 give an idea of the method,

screenshot at 50 helps with multiplication

screen shot at 48 give an idea of the method,

screenshot at 50 helps with multiplication

Original post by Sir Cumference

I may be wrong but I wouldn't expect to see a question like this in the final exam because the further maths dot product method is the quicker approach (cross product is even faster if you do further pure). A Level maths questions are normally designed so that on spec methods are the quickest and most efficient.

I am in complete agreement with you but the practice book that he is using is based on the new spec so there must be something with that question that we are missing

Original post by 14yalamanchilig

I am in complete agreement with you but the practice book that he is using is based on the new spec so there must be something with that question that we are missing

I posted a way to do it without further maths methods above. It just involves simple GCSE trig.

nah that method gave me 85.68

cause 1^2 +10^2+3^2=110

and 8^2+3^2+2^2=77

and theta is 68.6 from above

using |a||b|*sin(theta)=85.68

something is off

cause 1^2 +10^2+3^2=110

and 8^2+3^2+2^2=77

and theta is 68.6 from above

using |a||b|*sin(theta)=85.68

something is off

Original post by Sir Cumference

I posted a way to do it without further maths methods above. It just involves simple GCSE trig.

cool thanks for the help

Original post by 14yalamanchilig

i am bored so here it is...

screen shot at 48 give an idea of the method,

screenshot at 50 helps with multiplication

screen shot at 48 give an idea of the method,

screenshot at 50 helps with multiplication

I think that is a bit complicated for this course, but thanks anyway.

Original post by 14yalamanchilig

nah that method gave me 85.68

cause 1^2 +10^2+3^2=110

and 8^2+3^2+2^2=77

and theta is 68.6 from above

using |a||b|*sin(theta)=85.68

something is off

cause 1^2 +10^2+3^2=110

and 8^2+3^2+2^2=77

and theta is 68.6 from above

using |a||b|*sin(theta)=85.68

something is off

sorry theta is 110.35. i used cosine rule to find it.

(edited 4 years ago)

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