The Student Room Group
Reply 1
Bump.
Reply 2
Reply 3
Parallel - Direction vectors are multiplied by a factor
Perpendicular - Dot product of the 2 direction vectors = 0
Intersect - Solve simultaneous equations involving the 2 vector equations
Sakujo
How do I show that a pair of vectors are parallel/skewed/perpendicular/intersect? For parallel is it that their direction vectors are multiples of each other? Perpendicular, dot product= 0? Not sure on skewed or intersect.

Also, if someone could help me with question 4)ii) it would be much appreciated. C4 code is 4724.

http://www.ocr.org.uk/download/ms_08/ocr_16587_ms_08_l_gce_jun.pdf

http://www.ocr.org.uk/download/pp_08_jun/ocr_30882_pp_08_jun_l_gce_4724.pdf

Thanks in advance.:smile:


I think for skewed vectors you'd have to calculate the dot product and show that it's not 0 (i.e. the vectors are perpendicular to each other, as you quite rightly stated, since cos90=0) or an integer (i.e the directional vectors are parallel, arising by the fact that the cos0=1, thereby making the dot product the product of the two modulii). It may be that you are required to find the angle between the two directional vectors.

For an intersection I believe you'd have to show that their exists a point at which the vectors are equal. Set them equal to each other (with the constant for multiples of directional vectors included) and solve the arising equations simultaneously to show there exists such a point.

Please bear in mind I haven't done these in a while, so I may be completely off (or skewed)!
Reply 5
aspiringmathematician
I think for skewed vectors you'd have to calculate the dot product and show that it's not 0 (i.e. the vectors are perpendicular to each other, as you quite rightly stated, since cos90=0) or an integer (i.e the directional vectors are parallel, arising by the fact that the cos0=1, thereby making the dot product the product of the two modulii). It may be that you are required to find the angle between the two directional vectors.

For an intersection I believe you'd have to show that their exists a point at which the vectors are equal. Set them equal to each other (with the constant for multiples of directional vectors included) and solve the arising equations simultaneously to show there exists such a point.

Please bear in mind I haven't done these in a while, so I may be completely off (or skewed)!


Thank you.:smile:

Any ideas on the question I posted. The mark scheme is rubbish.
I've had it with these motherf##king Vectors on the Motherf##king X,Y Plane!
Reply 7
aspiringmathematician
I think for skewed vectors you'd have to calculate the dot product and show that it's not 0 (i.e. the vectors are perpendicular to each other, as you quite rightly stated, since cos90=0) or an integer (i.e the directional vectors are parallel, arising by the fact that the cos0=1, thereby making the dot product the product of the two modulii). It may be that you are required to find the angle between the two directional vectors.

For an intersection I believe you'd have to show that their exists a point at which the vectors are equal. Set them equal to each other (with the constant for multiples of directional vectors included) and solve the arising equations simultaneously to show there exists such a point.

Please bear in mind I haven't done these in a while, so I may be completely off (or skewed)!


Actually skew vectors are non-parallel vectors which don't intersect. To show that two vectors are skew, first show they aren't parallel (ie show that there exists no λ\lambda such that λa=b\lambda \mathbf{a}=\mathbf{b} for two vectors with directions a and b.

Then, you convert into a system of 3 simultaneous equations in two variables, and show that if you take two and solve for that the resulting values aren't a solution for the third.

The dot product of two skew vectors can be 0 (if they're perpendicular but don't intersect), and it can be an integer too (since a.b=|a||b|cos theta and if one of |a| or |b| is even and theta=π/3 then a.b is an integer but a and b could still be skew).
Good stuff, thanks for giving a correct answer!
Reply 9
Could anyone do the question that the OP originally posted ?

I had a go but could not get anywhere with it.
Reply 10
cyborg
Could anyone do the question that the OP originally posted ?

I had a go but could not get anywhere with it.


Same.
Reply 11
The markscheme does not help - consider their r and (-2,1,1) ??
Tua_the_samoan
I've had it with these motherf##king Vectors on the Motherf##king X,Y Plane!

so no problems with them in the Z plane?
apocalipse117
so no problems with them in the Z plane?


ha ha, the third dimension is too much for me lol

You doing edexcel c4?? if yes, you got any good websites or notes or anything please

Thanks
wat does dot product mean
Reply 15
Sakujo
How do I show that a pair of vectors are parallel/skewed/perpendicular/intersect? For parallel is it that their direction vectors are multiples of each other? Perpendicular, dot product= 0? Not sure on skewed or intersect.

Also, if someone could help me with question 4)ii) it would be much appreciated. C4 code is 4724.

http://www.ocr.org.uk/download/ms_08/ocr_16587_ms_08_l_gce_jun.pdf

http://www.ocr.org.uk/download/pp_08_jun/ocr_30882_pp_08_jun_l_gce_4724.pdf

Thanks in advance.:smile:


could the op (or any smart person) talk me through that question?
i don't understand how you find 1/6
Reply 16
miaow.
could the op (or any smart person) talk me through that question?
i don't understand how you find 1/6




Someone else PMed me the solution

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