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Mathematics 10 Online
OpenStudy (anonymous):

the rectangular box has a 2x2 base and height of 1. cosine of angle between AC and AB is...

OpenStudy (anonymous):

OpenStudy (anonymous):

hints please, trying to solve on my own...

OpenStudy (mertsj):

We need to see the rest of the problem

OpenStudy (anonymous):

i posted attachment

OpenStudy (mertsj):

I know and it's not on that either

OpenStudy (anonymous):

yes it is

OpenStudy (anonymous):

you sure? i'm pretty sure it's solveable, just can't remember how...i believe the angle between the vector and AB (i'll label x axis) would be roh(?)

OpenStudy (anonymous):

divide the problem into two right triangles

OpenStudy (anonymous):

elica85. First, project the line down onto the base square of the triangle. Find the hypotenuse of the triangle that is formed by doing this. This is our adjacent quantity. We know the opposite quantity. Now we can use the tangent function.

OpenStudy (anonymous):

whoops. project the line down onto the base rectangle of the prism! sorry

OpenStudy (anonymous):

or use dot product of AC with AB

OpenStudy (mertsj):

The cosine of the angle is 2/3

OpenStudy (anonymous):

ok, the dot product gives me the cosine? anything else i need to do after? and why dot produvct?

OpenStudy (anonymous):

because it gives you the cos of the angle, which is what you are trying to solve for

OpenStudy (anonymous):

|dw:1327112719359:dw|the dot product of A and B is C

OpenStudy (anonymous):

so,\[A*B=|AC||AB|\cos \theta\]

OpenStudy (anonymous):

let me know if you need more help from here

OpenStudy (anonymous):

Sorry I should have wrote \[AC*AB=|AC||AB|\cos \theta\]

OpenStudy (anonymous):

thx, still trying to understand...first how do i get AC...it would be sqrt(1^2+?^2)

OpenStudy (anonymous):

no

OpenStudy (anonymous):

o, sqrt(1+8)=3

OpenStudy (anonymous):

so AC=3

OpenStudy (anonymous):

AC=<2,2,1> so |AC|=(2^2+2^2+1^2)^1/2

OpenStudy (anonymous):

ok...i saw that dot product formula quite a few times but i don't get it. <AB>dot<AC>=|AB||AC|...and multiplied by cos...

OpenStudy (anonymous):

yeah, exactly. |AB||AC|cosa, where a i the angle between vectors AB and AC

OpenStudy (anonymous):

do you remember how to calculate the dot product part?

OpenStudy (anonymous):

yes. mult component wise and then add

OpenStudy (anonymous):

yep :)

OpenStudy (anonymous):

ok so have that on the left side...and then solve for cosa

OpenStudy (anonymous):

exactly

OpenStudy (anonymous):

ok ok. so use sina if the question asks to find the sine of the angle?

OpenStudy (anonymous):

no..the dot product only gives you cosines

OpenStudy (anonymous):

so if i'm asked for sine...?

OpenStudy (anonymous):

then you need to use another relation. if you have corresponding angles for example, you can relate the cos of one angle to the sin of another.

OpenStudy (anonymous):

ok, thx you closed the gap a lot for me just from this problem

OpenStudy (anonymous):

or the question might involve the cross-product of two vectors which involve the sin of the angle between the two vectors..depends on the problem

OpenStudy (anonymous):

np

OpenStudy (anonymous):

i get .84 but the closest out of the choices i have is 2/sqrt(5)...or none of the above

OpenStudy (anonymous):

I get cos a = 2/3

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