Can someone start me off on these and please explain step by step how you got everything because I'm having difficulty understanding the entire concept.
Find the projection of u onto v and calculate its magnitude.
a) u = (2,5) v = (6,4)
b) u = (3, 6, -2) v = (-4, 3, 8)
2) If u and v are non zero vectors, but Projection (u unto v) = 0, what conclusion can be drawn?
Would that also mean Projection (v unto u) = 0?
3) Find the projection of PQ onto each of the coordinate axes, where point (2,3,5) and Q is the point (-1,2,5).
It is said simply here
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/dotprod/dotprod.html
If the projection is zero, the dot product is zero, and thelines are perpendicular.
a) To find the projection of vector u onto vector v, you can use the following formula:
projv(u) = (u·v / ||v||^2) * v
where · denotes the dot product and ||v|| indicates the magnitude of vector v.
Step-by-step solution:
1. Calculate the dot product of u and v: (2*6) + (5*4) = 12 + 20 = 32.
2. Calculate the magnitude of v: ||v|| = sqrt(6^2 + 4^2) = sqrt(36 + 16) = sqrt(52) = 2*sqrt(13).
3. Plug the values into the formula: projv(u) = (32 / (2*sqrt(13))^2) * (6, 4)
= (32 / 52) * (6, 4)
= (16/13) * (6, 4)
= (96/13, 64/13).
To calculate the magnitude of the projected vector, simply find its length:
||projv(u)|| = sqrt((96/13)^2 + (64/13)^2)
= sqrt(9216/169 + 4096/169)
= sqrt(13312/169)
= sqrt(78.688)
≈ 8.86.
b) Using the same formula as above:
Step-by-step solution:
1. Calculate the dot product of u and v: (3*-4) + (6*3) + (-2*8) = -12 + 18 - 16 = -10.
2. Calculate the magnitude of v: ||v|| = sqrt((-4)^2 + 3^2 + 8^2) = sqrt(16 + 9 + 64) = sqrt(89).
3. Plug the values into the formula: projv(u) = (-10 / sqrt(89)^2) * (-4, 3, 8)
= (-10 / 89) * (-4, 3, 8)
= (40/89, -30/89, -80/89).
To calculate the magnitude of the projected vector, find its length:
||projv(u)|| = sqrt((40/89)^2 + (-30/89)^2 + (-80/89)^2)
= sqrt(1600/7921 + 900/7921 + 6400/7921)
= sqrt(8900/7921)
≈ 1.88.
2) If Projection (u unto v) = 0 and u and v are non-zero vectors, it implies that u and v are orthogonal or perpendicular to each other. This means they form a right angle between them. In this case, it does not necessarily mean that Projection (v unto u) = 0. The projection onto v could be zero while the projection onto u might not be zero. The result would depend on the angle between u and v.
3) To find the projection of vector PQ onto each of the coordinate axes (x, y, and z), you need to calculate the dot product of PQ with each of the coordinate axes unit vectors (i, j, and k).
Step-by-step solution:
1. Subtract the coordinates of point Q from point P to get vector PQ:
PQ = (2, 3, 5) - (-1, 2, 5) = (2+1, 3-2, 5-5) = (3, 1, 0).
2. Calculate the dot product of PQ with each of the coordinate axes unit vectors:
projx(PQ) = (3*1) + (1*0) + (0*0) = 3,
projy(PQ) = (3*0) + (1*1) + (0*0) = 1,
projz(PQ) = (3*0) + (1*0) + (0*0) = 0.
Therefore, the projection of PQ onto the x-axis is 3, onto the y-axis is 1, and onto the z-axis is 0.