(mississauga, ontario, Canada)
i have to find the scalar and vector projection of a=i-j+k and b=2i-j-2k
and i got:
Vector proj = (1/3)(i-j+k) = i/3 + j/3 + k/3
scalar proj = (1/9)(2i-j-2k) = 2i/9 - j/9 - 2k/9
is this correct?
I am given a graph of f’(x) and am told to graph f”(x) and f(x)
How would i do this question?
Determine the distance between r=(-2,1,0)+s(1,-1,1) and r=(0,1,0)+t(1,1,2)
My common perpendicular was: 3i+j-2k and the vector joining the two points (2,0,0) and then i did this:(-2,1,0)-(0,1,0)
Would this be correct?
determine weather r=(16,11,-24) can be written as a linear combination of p=(-2,3,4) and q=(4,1,-6). I got that is cannot. Would that be correct? I can also show my work if anyone wants.
X and y are vectors of magnitude 1 and 2. With an angle of 120 degrees between them. Determine 3x+2y and the direction. I used the cosine law but subbed in 3 and 2 and cos 120. And i got sqrt37 as my overall answer and i found the direction by taking
sinA/4=sin120/sqrt37 and i sin inversed and got my final answer of 34 degrees. Would that be correct?
Two forces of 40N and 50N act at an angle of 60degrees determine resultant and equilibrant this is what i did: 180-60=120 use cosine:
40^2+50^2-2(40)(50)cos120=78.10 ß resultant
(40)(sin120)/78.10= 26.3 degrees ßequilibriant?
Would this be correct?
An airplan is heading due north at 1000km/h when it encounters wind from the east at 100km/h. Determine resultant velocity
I drew a diagram and then: w=100 v=1000 resultant velocity:w+v
I used the cosine formula with the angle 90 degrees because that is what the angle east makes with north. And i got 1005. Then i used sina=100(sin90)/1005= 5.71 degrees at 1005km/h as my final answer..is this correct?
A canoeist wants to cross a 200m river to get to a campsite directly across from teh starting point. The canoeist can paddle at 2.5m/s and the current has a speed of 1.2m/s
How far downstream would the canoeist land if headed directly across the river?
In what direction should the canoeist head in order to arrive directly across from the starting point.
What i did was that i drew a diagram and made a triangle. And then: 1.2/2.5=d/200 d=96, the canoeist will land 96m downstream.
And then sina=(1.2/2.5) a=28.68degrees
(v+w)^2= 6.25-1.44 =2.19
The canoeist has to travel upstream 28.68 degrees at the speed of 2.19 to arrive directly across from the starting point. Would this be correct?
Calculate the area of a triangle with the vertices: A(-1,3,5) B(2,1,3) C(-1,1,4)
What i did was:
AB=sqrt((1-2)^2+(3-1)^2+(5-3)^2)= sqrt17 =a
BC=sqrt((2+1)^2 + (1-1)^2 + (3-4)^2 )= sqrt10=c
=3.49 is this correct?