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C4 Exam style paper Q6

Hi,

I'm currently stuck on Q6 of the Examination style paper....

Relative to a fixed origin O, the three points A, B and C have position vectors a=2i+6j-5k, b=4i+10j-9k and c=8j-7k respectively.

a) find the angle ABC giving your answer to the nearest degree.

I got the answer as 35 which is right :smile:

b) the point D lies on the line through A and C and OD is perpendicular to AC. Find the position vector of D.

c)Hence find the exact value of the area of triangle OAC


thanks :smile:

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Original post by JEvans1234
b) the point D lies on the line through A and C and OD is perpendicular to AC. Find the position vector of D.


When two vectors are perpendicular (orthogonal might be a better word to use here), their scalar product is equal to 0.

Which two direction vectors are orthogonal to each another?
Reply 2
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JEvans1234
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Original post by Khallil
When two vectors are perpendicular (orthogonal might be a better word to use here), their scalar product is equal to 0.

Which two direction vectors are orthogonal to each another?


OD and AC..... so you would do OD . AC = 0 ????
Original post by JEvans1234
OD and AC..... so you would do OD . AC = 0 ????


You tell me.
Reply 4
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JEvans1234
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Original post by Khallil
You tell me.



yes im pretty sure you would.... when you have two lines that are perpendicular the dot product is zero and the two lines in question are OD and AC
Original post by JEvans1234
yes im pretty sure you would.... when you have two lines that are perpendicular the dot product is zero and the two lines in question are OD and AC


Have a go then!
Reply 6
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JEvans1234
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Original post by Khallil
Have a go then!


ive tried the dot product and im still not getting the answer which is 5i+3j-2k..... the question is 6 marks so i doubt its just the product rule......
Reply 7
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the bear
20
can we see your equation for line AC ?
Reply 8
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JEvans1234
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Original post by the bear
can we see your equation for line AC ?


AC--> = C - A = -2i + 2j + -2k
Reply 9
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the bear
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Original post by JEvans1234
AC--> = C - A = -2i + 2j + -2k


that is part of the equation... not all of it
Reply 10
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JEvans1234
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Original post by the bear
that is part of the equation... not all of it


ohhh...... whats the rest of it then???
Reply 11
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the bear
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it should be in the form r = p + λq
Reply 12
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JEvans1234
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Original post by the bear
it should be in the form r = p + λq


so it would be 2i + 6j - 5k + lambda(-2i +2j -2k)??
Original post by JEvans1234
so it would be 2i + 6j - 5k + lambda(-2i +2j -2k)??


:yep:
Reply 14
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the bear
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Original post by JEvans1234
so it would be 2i + 6j - 5k + lambda(-2i +2j -2k)??


yes... although you have the option of cancelling down the direction part...
Original post by JEvans1234
so it would be 2i + 6j - 5k + lambda(-2i +2j -2k)??


Now see if you can set up an equation for OD in terms of a⃗\vec{\mathbf{a}} and c⃗\vec{\mathbf{c}}.
Reply 16
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JEvans1234
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Original post by Khallil
Now see if you can set up an equation for OD in terms of a⃗\vec{\mathbf{a}} and c⃗\vec{\mathbf{c}}.


so OD = OA + AD and OD = OC + CD
Original post by JEvans1234
so OD = OA + AD and OD = OC + CD


They're both equally valid, but let us go with the first one.

How can you express AD in terms of a⃗\vec{\mathbf{a}} and c⃗\vec{\mathbf{c}} ?

The point D lies on the line through A and C
Reply 18
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JEvans1234
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Original post by Khallil
They're both equally valid, but let us go with the first one.

How can you express AD in terms of a⃗\vec{\mathbf{a}} and c⃗\vec{\mathbf{c}} ?


would it be.... AD = AC - CD ????
Original post by JEvans1234
would it be.... AD = AC - CD ????


Think about scalar multiples of the direction vectors of parallel lines.

d⃗=OA⃗+AD⃗=a⃗+λ(...)\begin{aligned} \vec{d} & = \vec{OA} + \vec{AD} \\ & = \vec{a} + \lambda \left( ... \right) \end{aligned}

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