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Mathematics: Post your doubts here!

Thought blocker

Thought blocker

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sandman za said:
could someone pleeease explain the vector question 7 (i) and (ii) from May June 2014 paper 13???

got the marking scheme but i dont understand :0
Click to expand...
7i)
To find angle BAC, you will need sides, take it either AB, AC or BA, CA
Lets chose AB and AC. So now, we need vectors AB and AC, hence
We need to do b - a for AB and c - a for AC.
AB = (4, -2, 4) And AC = (0, 3, 4)
Now the second step we need to do is dot product of 2 vectors AB and AC.
Thats AB.AC = (4*0, -2*3, 4*4) = 0 - 6 + 16 = 10
Opening vectors,
Root(ABsquared) * Root(ACsquared) * cosBAC = 10
Root(36) * Root(25) * cosBAC = 10
BAC = cos inverse 1/3
But RoOkaYya G if we do it with BA and CA its not possible...Why? o_O Rutzaba

7ii)
RoOkaYya G help :p
 
RoOkaYya G

RoOkaYya G

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Thought blocker said:
7i)
To find angle BAC, you will need sides, take it either AB, AC or BA, CA
Lets chose AB and AC. So now, we need vectors AB and AC, hence
We need to do b - a for AB and c - a for AC.
AB = (4, -2, 4) And AC = (0, 3, 4)
Now the second step we need to do is dot product of 2 vectors AB and AC.
Thats AB.AC = (4*0, -2*3, 4*4) = 0 - 6 + 16 = 10
Opening vectors,
Root(ABsquared) * Root(ACsquared) * cosBAC = 10
Root(36) * Root(25) * cosBAC = 10
BAC = cos inverse 1/3
But RoOkaYya G if we do it with BA and CA its not possible...Why? o_O Rutzaba

7ii)
RoOkaYya G help :p
Click to expand...
it wont work of course -___- coz u r searching for the angle A. which is between AB and AC....so u need to take A as starting point. its vectors so the direction counts hell lot. so u need to strt from A. so ull be needed vector directions AC and AB.thts how ive learnt it. ask others if theres other explanations.
KIDDO!! HOW DOES IT BECOME 4 WHEN U DO b-a FOR VECTOR AB? :eek: im getting -2 :(
for the i position ---> (4,-2,4)
7ii) use formula for area of triangle 0.5*AB*sinC
So in this case u shld AC and AB using distance formula
use the answer in part (i) and derive the sin BAC from it using identities. or u can even use right angled triangle n derive tht.
then u replace in area of triangle formula.
 
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sandman za

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Thought blocker said:
7i)
Opening vectors,
Root(ABsquared) * Root(ACsquared) * cosBAC = 10
Root(36) * Root(25) * cosBAC = 10
BAC = cos inverse 1/3
But RoOkaYya G if we do it with BA and CA its not possible...Why? o_O Rutzaba

7ii)
RoOkaYya G help :p
Click to expand...

why do we root and square here?

and the part (ii), please could you write out the deriving part? i really find this question difficult :/
 
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sandman za

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RoOkaYya G said:
coz its their magnitude i.e distance
Click to expand...

i have vector AB (-2, -2, 4) which has magnitude 2*root(6)

vector AC (0, 3, 4) has magnitude 5

a.b = 0 -6 + 16 = 10

so we have 10 = 2*root(6) * 5 * cosBAC

cosBAC = root(6) / 6

cos inverse of that gives 65.9, which is not the same as cos inverse of 1/3 which is 70.5

have i made a mistake somewhere here?
 
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