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Mathematics: Post your doubts here!

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2pac

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http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_31.pdf
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_ms_31.pdf
Can someone please help me with the third part of q7.I know we're supposed to use the cross product for such type of questions but I can't seem to understand which directional vectors to take.The first one will be the directional vector of AB but what about the other one.How will we find the directional vector of OAB?
Please help me.
 
Aarjit

Aarjit

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2pac said:
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_31.pdf
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_ms_31.pdf
Can someone please help me with the third part of q7.I know we're supposed to use the cross product for such type of questions but I can't seem to understand which directional vectors to take.The first one will be the directional vector of AB but what about the other one.How will we find the directional vector of OAB?
Please help me.
Click to expand...

You have: OP = (2/3 5/3 7/3) [from (ii)]
Since AB and OP are orthogonal, and the plane you're solving for contains AB, OP must be orthogonal to the plane as well!
Take out the scalar factor (1/3) to find the normal vector n ;
>> n = (2 5 7)
you have A (1 2 2);
now, the requisite equation for the plane:

a.n=r.n
>> (1 2 2).(2 5 7) = (x y z).(2 5 7)
>> 2 +10+14 = 2x + 5y +7z
>> 2x + 5y +7z = 26

Q.E.D
 
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2pac

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Aarjit said:
You have: OP = (2/3 5/3 7/3) [from (ii)]
Since AB and OP are orthogonal, and the plane you're solving for contains AB, OP must be orthogonal to the plane as well!
Take out the scalar factor (1/3) to find the normal vector n ;
>> n = (2 5 7)
you have A (1 2 2);
now, the requisite equation for the plane:

a.n=r.n
>> (1 2 2).(2 5 7) = (x y z).(2 5 7)
>> 2 +10+14 = 2x + 5y +7z
>> 2x + 5y +7z = 26

Q.E.D
Click to expand...
okay I didn't understand much probably because I am used to equating two directional vectors and solving them with the cross product.I then take the position vectors of the line on which the plane contains and then equate the result of cross product subsequently to find 'd'.
when the question says its perpendicular to OAB,why do we consider OP at all?
 
Mobeen

Mobeen

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2pac said:
Recheck your answer,you might have made a silly sign mistake.That's what I think though.
Click to expand...
i dont know man ! did it over and over again ! .. can you do this and just tell me if you get the right answer ?
 
johnsth

johnsth

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i
Keitak said:
Thanks for replying!
when you make X the subject does the answer you get have to be within the new range or the one specified in the question?
Click to expand...
if u r talkin about alpha it can come anything usually in 60's the theta in 2nd part has to be in the specified range in the question
 
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2pac

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Mobeen said:
i dont know man ! did it over and over again ! .. can you do this and just tell me if you get the right answer ?
Click to expand...
okay i got it.
i hope u know how to the first step.there u see that the denominator can be factorized to (1-3tan^2theeta).before factorizing that cross multiply both sides and solve.after uve simplified the numerator,you'll get the answer.
 
johnsth

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