Vectors help

https://gyazo.com/46db641a509c974ff607c8af742272cd
I'm not really sure on these 2 questions, I tried using a.b/moda x mod b= cos theta equation to find equation where the 2 position vectors meet, I got 40.2(3sf) while it's asking for an exact value

for b, I tried using 1/2absinc for the area so I tried doing root 6 x root 14 x 40.2.. and I can't seem to get it in the form expected in the question

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Reply 1

CaptainDuckie

Study Forum Helper

For a, you can find one length of the other side

Then use cosine rule, I think

Reply 2

HelloZello

Original post by CaptainDuckie

For a, you can find one length of the other side

Then use cosine rule, I think

oh right ok that makes sense i was thinking that but it didn't seem right thank you

Reply 3

MiladA

For a) It may be helpful to draw a sketch of angle OXY. Remember it is the length AWAY from the angle you need. So calculate XO and XY and use formula you mentioned.

For b) use (1/2)*(a)*(b)*sinC

Reply 4

CaptainDuckie

Study Forum Helper

Original post by HelloZello

oh right ok that makes sense i was thinking that but it didn't seem right thank you

What did you get?

Reply 5

HelloZello

Original post by CaptainDuckie

What did you get?

am completing the last questions am gonna come bk to that 1 later hopefully- what did u get so i can compare my answers

Reply 6

CaptainDuckie

Study Forum Helper

Original post by HelloZello

am completing the last questions am gonna come bk to that 1 later hopefully- what did u get so i can compare my answers

.... I’m not supposed to tell you answers

Reply 7

DFranklin

Original post by HelloZello

This is the right approach; you can express the length of the vectors exactly (as the square root of an integer).

Reply 8

HelloZello

Original post by DFranklin

This is the right approach; you can express the length of the vectors exactly (as the square root of an integer).

i thought they ment exact trig values so i was confused before

Reply 9

MiladA

Original post by HelloZello

i thought they ment exact trig values so i was confused before

The angle is correct. Leave it as cos(OXY) = .... as this is exact value. For part b) draw a triangle and calculate sin(OXY) (or use sin(OXY) = sqrt(1-cos(OXY)^2) and the area of a triangle formula.

Reply 10

HelloZello

Original post by MiladA

The angle is correct. Leave it as cos(OXY) = .... as this is exact value. For part b) draw a triangle and calculate sin(OXY) (or use sin(OXY) = sqrt(1-cos(OXY)^2) and the area of a triangle formula.

i diid 1/2absinc and I did 1/2 x root 14 x root 6 x sin(40.2) as thats the angle in between and I got 2.957..=2.96(3sf) and I don't know how to get this into exact form...

Reply 11

davros

Original post by HelloZello

i diid 1/2absinc and I did 1/2 x root 14 x root 6 x sin(40.2) as thats the angle in between and I got 2.957..=2.96(3sf) and I don't know how to get this into exact form...

As per above advice, leave cos C as an exact value, then calculate sin C from this, so it too will be an exact value.

Do you have an answer to check against?

Reply 12

HelloZello

Original post by davros

As per above advice, leave cos C as an exact value, then calculate sin C from this, so it too will be an exact value.

Do you have an answer to check against?

dont think i learnt that yet- does that relate to quadrants perhaps, I'm not really sure how to do that sorry, does it involve complementary angles where they are 90 degrees apart? spent so much time on the 2 markers but I done the last questions involving shortest distances

Reply 13

davros

Original post by HelloZello

You've never come across

\sin^2 A + \cos^2 A = 1

before ?

Reply 14

HelloZello

Original post by davros

You've never come across

\sin^2 A + \cos^2 A = 1

before ?

i have sir oooh i get it now its night lol my brain is a bit fried

(edited 3 years ago)

Reply 15

davros

Original post by HelloZello

i have sir oooh i get it now its night lol my brain is a bit fried

So what's the problem? You worked out cos C (or whatever angle you called it) as an exact value in the first part, so now you want to know what sin C is for the area formula...

EDIT: just seen you've updated your post :smile:

Reply 16

HelloZello

Original post by davros

thank u haha

(edited 3 years ago)

Reply 17

davros

Original post by HelloZello

thank u haha

So have you now got an answer in the required form? (You shouldn't be working out any angles or taking the sin or cos of anything :smile:

)

Reply 18

HelloZello

Original post by davros

So have you now got an answer in the required form? (You shouldn't be working out any angles or taking the sin or cos of anything :smile:

)

i thought i knew the answer but i forgot how to do it- we havent learn this in school yet so yeah i keep having an idea and forgetting how to do it, this question is only 2 marks but still :c