Vectors

[Lee_Truss]

Hey there,

Could someone take a look at this question,

The angle better vectors i + j and 2i + j + tk is pi/4. Find the possible values of t.

Ive been at it for a while now and cant seem to get the right answer.

[kedaman]

cos a =(V1 dot V2)/(|V1||V2|)
solve V1.z

[Lee_Truss]

Yad think, ive been trying that and no luck.

[Guv]

Your notation is unfamiliiar to me. I assume that the two vectors in mathematical instead of engineering noation are the following.

(1, 1, 0) & (2, 1, t)

The dot (or scalar product) of those vectors is 3, which is not a function of t.

The cosine of the angle is 3 / SquareRoot(10)

Hence the angle is about 18.4349 degrees, not matter what value you assign to t.

[kedaman]

i j and k is called the unit vectors and defines a coordinate system.
(1, 1, 0) dot (2, 1, t) is a function of t, i'm just too lazy to do it.
well i guess i have nothing else to do
((1*t-0*1) - (1*t-0*2) + (1*1-1*2))/(sqr(1^2+1^2)sqr(2^2+1^2+t^2))=cos(pi/4)
the rest is yours

[sql_lall]

((1*t-0*1) - (1*t-0*2) + (1*1-1*2))/(sqr(1^2+1^2)sqr(2^2+1^2+t^2))=cos(pi/4)

IS THE SAME AS:

(t-t-1)/(sqr(2)*sqr(5+t^2)) = cos(pi/4)
- I just simplified stuff like 1^2, 1*t etc.

IS THE SAME IS

-1/(sqr(2)*sqr(5+t^2)) = cos(pi/4)

=> -1/cos(pi/4)=sqr(2)*sqr(5+t^2)
=> -1/(cos(pi/4) * sqr(2)) = sqr(5+t^2)
=> sqr( 1/( (cos(pi/4) * sqr(2)) ^2) -5)=t

[kedaman]

just call me lazy (sorry guv, yeah you were right but then again the t was in |V2|)

[Lee_Truss]

Hey there,

I got the answers (+ and - 2), I left my calculator in degrees and not radians, thanks anyway.

[DerFarm]

Your welcome. Glad to be of service. Call back anytime, Kovan
and Guv are waiting.

[kovan]

how did i get involved in this?