two vectors are defined as follows: ➡️AC = (-2 -1) ➡️AB = (-1, -4) find the value of cos(ACB) in its simpliest form

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Original post by mqb2766

Have you sketched it? If theyre asking for cos(...) it would suggest using the cos rule?

i literally have no clue how to do it tho 😭 we haven’t done vectors yet at school i’m just trying to learn it before as we start it after christmas

(edited 10 months ago)

Original post by annabel whittle

i literally have no clue how to do it tho 😭 we haven’t done vectors yet at school i’m just trying to learn it before as we start it after christmas

Maybe just wait until you do them at school?

But im sure you can sketch the two given vectors and use pythagoras to get their lengths. Then get BC and its length using the usual difference of the given vectors and use the cos rule on the triangle.

Original post by mqb2766

Maybe just wait until you do them at school?

But im sure you can sketch the two given vectors and use pythagoras to get their lengths. Then get BC and its length using the usual difference of the given vectors and use the cos rule on the triangle.

But im sure you can sketch the two given vectors and use pythagoras to get their lengths. Then get BC and its length using the usual difference of the given vectors and use the cos rule on the triangle.

sorry if this sounds dumb but i thought u only did that when it has the 2 lines around it like the modulus

Original post by annabel whittle

sorry if this sounds dumb but i thought u only did that when it has the 2 lines around it like the modulus

|AC| is the modulus or length of the vector and you use pythagoras to calculate it.

Original post by mqb2766

|AC| is the modulus or length of the vector and you use pythagoras to calculate it.

but it’s not written like that tho 😭 it’s just AC with this ➡️ above it

Original post by annabel whittle

but it’s not written like that tho 😭 it’s just AC with this ➡️ above it

This is basics and it would be worth waiting.

But AC = (-2,-1) means its x component is -2 and its y component is -1, and assuming A is at the origin, you use pythagoras to get

|AC| = sqrt((-2)^2 + (-1)^2) = sqrt(5)

(edited 10 months ago)

Original post by mqb2766

This is basics and it would be worth waiting.

But AC = (-2,-1) means its x component is -2 and its y component is -1, and assuming A is at the origin, you use pythagoras to get

|AC| = sqrt((-2)^2 + (-1)^2) = sqrt(5)

But AC = (-2,-1) means its x component is -2 and its y component is -1, and assuming A is at the origin, you use pythagoras to get

|AC| = sqrt((-2)^2 + (-1)^2) = sqrt(5)

ohh okay that makes sm more sense now! thank u should i just do the cosine rule now on the triangle of do i have to do anything else?

Original post by annabel whittle

ohh okay that makes sm more sense now! thank u should i just do the cosine rule now on the triangle of do i have to do anything else?

Once youve calculated the lengths of the 3 sides (using pythagoras/modulus on the vectors), then plugging the values into the cos rule gives cos(...).

Original post by mqb2766

Once youve calculated the lengths of the 3 sides (using pythagoras/modulus on the vectors), then plugging the values into the cos rule gives cos(...).

okay thank u sm for helping me!

Original post by annabel whittle

okay thank u sm for helping me!

Note an alternative would be to split <ACB into two with a horizontal line and do simple trig on each part (using the right triangles defined by the vector coordinates) and then combine. Probably a bit faster.

Original post by mqb2766

Note an alternative would be to split <ACB into two with a horizontal line and do simple trig on each part (using the right triangles defined by the vector coordinates) and then combine. Probably a bit faster.

hiya again i did the quesion and got this as my answer do you mind checking it for me?

Original post by annabel whittle

hiya again i did the quesion and got this as my answer do you mind checking it for me?

Image isnt displaying for me.

Original post by tonyiptony

So, another method that's much faster (if you happen to have looked up what a dot product is)...

Notice that $CA \cdot CB = |CA||CB|\cos (\angle ACB)$. Now plug in everything (be careful though, we don't know CB yet).

Notice that $CA \cdot CB = |CA||CB|\cos (\angle ACB)$. Now plug in everything (be careful though, we don't know CB yet).

Good method.

However, I doubt OP would learn dot product at the GCSE level.

Original post by Eimmanuel

Good method.

However, I doubt OP would learn dot product at the GCSE level.

However, I doubt OP would learn dot product at the GCSE level.

thank you !

Original post by mqb2766

Image isnt displaying for me.

oh that’s bad dyk any ways it would work for you

Original post by annabel whittle

oh that’s bad dyk any ways it would work for you

Maybe post the image on another site and link it.

Original post by mqb2766

This is basics and it would be worth waiting.

But AC = (-2,-1) means its x component is -2 and its y component is -1, and assuming A is at the origin, you use pythagoras to get

|AC| = sqrt((-2)^2 + (-1)^2) = sqrt(5)

But AC = (-2,-1) means its x component is -2 and its y component is -1, and assuming A is at the origin, you use pythagoras to get

|AC| = sqrt((-2)^2 + (-1)^2) = sqrt(5)

You don't need to assume A is the origin nor do you need to use vectors to get BC - the sketch gives it to you.

Original post by annabel whittle

hiya again i did the quesion and got this as my answer do you mind checking it for me?

Just write your answer

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