Find the scalar and vector projections of b onto a.
a = 2i - j + 6k, b = j + (1/2)k
compab =
projab =
Well, well, well, it seems we are dealing with some projections here. Let's get calculating!
To find the scalar projection of b onto a, we'll use the formula:
scalar projection of b onto a = |b| * cos(theta),
where |b| is the magnitude of b and theta is the angle between a and b.
Now, the magnitude of b is |b| = sqrt(j^2 + (1/2)^2 k^2) = sqrt(1 + 1/4) = sqrt(5/4) = sqrt(5)/2.
To find the angle theta, we can use the dot product:
a · b = |a| * |b| * cos(theta),
where |a| is the magnitude of a.
First, let's find the magnitude of a:
|a| = sqrt(2^2 + (-1)^2 + 6^2) = sqrt(4 + 1 + 36) = sqrt(41).
Now, let's find the dot product:
a · b = 2(0) - 1(1) + 6(1/2) = 0 - 1 + 3 = 2.
So, the equation becomes:
2 = sqrt(41) * (sqrt(5)/2) * cos(theta).
Dividing both sides by sqrt(41) * (sqrt(5)/2) gives us:
cos(theta) = 2 / (sqrt(41) * (sqrt(5)/2)).
cos(theta) = 4 / (sqrt(205)).
Now, to find the scalar projection:
scalar projection of b onto a = |b| * cos(theta)
= (sqrt(5)/2) * (4 / (sqrt(205)))
= 2 * (4 / (sqrt(205)))
= 8 / (sqrt(205)).
So, the scalar projection of b onto a is 8 / (sqrt(205)).
Now, let's move on to finding the vector projection of b onto a.
The formula for vector projection is:
vector projection of b onto a = (scalar projection of b onto a) * (unit vector of a)
To find the unit vector of a, we divide a by its magnitude:
|a| = sqrt(41), so
unit vector of a = (2i - j + 6k) / sqrt(41)
= (2 / sqrt(41))i - (1 / sqrt(41))j + (6 / sqrt(41))k.
And now, finally, we calculate the vector projection:
vector projection of b onto a = (8 / (sqrt(205))) * [(2 / sqrt(41))i - (1 / sqrt(41))j + (6 / sqrt(41))k]
= (16 / (sqrt(41)sqrt(205)))i - (8 / (sqrt(41)sqrt(205)))j + (48 / (sqrt(41)sqrt(205)))k.
So, the vector projection of b onto a is (16 / (sqrt(41)sqrt(205)))i - (8 / (sqrt(41)sqrt(205)))j + (48 / (sqrt(41)sqrt(205)))k.
I hope that didn't project too much confusion onto you. Have a laugh and keep on calculating!
To find the scalar and vector projections of b onto a, you can follow these steps:
Step 1: Calculate the dot product of vectors a and b.
Step 2: Calculate the magnitude of vector a.
Step 3: Calculate the scalar projection (compab) using the formula: compab = (a • b) / |a|.
Step 4: Calculate the vector projection (projab) using the formula: projab = (compab/|a|) * a.
Let's calculate these values step by step:
Step 1: Calculate the dot product of vectors a and b:
a • b = (2)(0) + (-1)(1) + (6)(1/2) = -1 + 3 = 2
Step 2: Calculate the magnitude of vector a:
|a| = sqrt((2)^2 + (-1)^2 + (6)^2) = sqrt(4 + 1 + 36) = sqrt(41)
Step 3: Calculate the scalar projection (compab):
compab = (a • b) / |a| = 2 / sqrt(41)
Step 4: Calculate the vector projection (projab):
projab = (compab/|a|) * a = (2 / sqrt(41)) * (2i - j + 6k) = (4i - 2j + 12k) / sqrt(41)
Therefore, the scalar projection (compab) of b onto a is 2 / sqrt(41), and the vector projection (projab) of b onto a is (4i - 2j + 12k) / sqrt(41).
Did you look at the solutions I gave you to
http://www.jiskha.com/display.cgi?id=1270516096
For this one ..
a = [2,-1,6] and b = [0,1,1/2]
follow the same steps.