A level maths vector question help asap

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TSRUser1231
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#1
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#1
In triangle ABC, AB = -3i+6j and AC=10i-2j

1.)
a.) find the size of angle BAC, in degrees to 1 decimal place.
b.) Find the exact value of the area of ABC

Any help will be appreciated!!
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Kevin De Bruyne
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#2
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#2
(Original post by TSRUser1231)
In triangle ABC, AB = -3i+6j and AC=10i-2j

1.)
a.) find the size of angle BAC, in degrees to 1 decimal place.
b.) Find the exact value of the area of ABC

Any help will be appreciated!!
What techniques do you know for finding the angle between two vectors?
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RickHendricks
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#3
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#3
(Original post by TSRUser1231)
In triangle ABC, AB = -3i+6j and AC=10i-2j

1.)
a.) find the size of angle BAC, in degrees to 1 decimal place.
b.) Find the exact value of the area of ABC

Any help will be appreciated!!
Not sure which question:

a. Best that you sketch  ABC on a  \mathbf{i} and  \mathbf{j} plane.

The best way to work out the angle would be to know the magnitude of  AB ,  AC and  BC

Once you know the magnitude, use Cosine rule to work out the angle made from BAC.

b. Once you know the angle and the magnitude of the 2 lines making the angle, use the formuala  A = \displaystyle\frac{1}{2} \cdot a \cdot b \cdot Sin(c)
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himmz
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#4
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#4
PART A
1. Work out vector BC. BC = AC - AB. Vector BC ends up being 13i - 8j
2. Work out the length of each vector using pythagoras. AB = sqrt( 3^2 6^2 ). AB = 3sqrt(5), AC = 2sqrt(26), BC = sqrt(233)
3. Use the cosine rule to work out angle BAC. A = Cos^-1 [( b^2 c^2 -a^2 ) / 2bc]. A ends up being 127.9 degrees.
PART B
1. Use the Area = 0.5abSinC rule. Angle C is A from the previous part.
Area ends up being 27 units^2.
Last edited by himmz; 1 year ago
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davros
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#5
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#5
(Original post by himmz)
PART A
1. Work out vector BC. BC = AC - AB. Vector BC ends up being 13i - 8j
2. Work out the length of each vector using pythagoras. AB = sqrt( 3^2 6^2 ). AB = 3sqrt(5), AC = 2sqrt(26), BC = sqrt(233)
3. Use the cosine rule to work out angle BAC. A = Cos^-1 [( b^2 c^2 -a^2 ) / 2bc]. A ends up being 127.9 degrees.
PART B
1. Use the Area = 0.5abSinC rule. Angle C is A from the previous part.
Area ends up being 27 units^2.
You do realise that you're replying to a thread that's 2 years old??
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getthruschool
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#6
let them, it helps future students - i'm here now aren't i LMAO
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davros
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#7
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#7
(Original post by getthruschool)
let them, it helps future students - i'm here now aren't i LMAO
It's against forum rules to post solutions - see the guidelines in the sticky post at the top of the maths forum. And there's no evidence that the OP ever came back or attempted the question themselves
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dan bilzerian
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#8
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#8
it actually did, thank you very much.
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