1)let w=[3;4] and u=[1;2]
a) find the projection p of u onto w. I found this to be p=[1.32;1.76]
b) find a scalar k for which the vector kp has a norm that is equal to one. k=?
d)find a vector that is orthogonal to span{w} s=?
how do I do this
#1 a unit vector in the direction of w is
z=[3/5;4/5]
p = u.z = 11/5 z = [33/25;44/25] = [1.32;1.76]
you are correct
|kp| = k|p| = 55/25 k
So, k = 25/55
As for the span, that is usually defined for a set of vectors. span{w} would just be w. So, any vector orthogonal to w will work.
could you explain how you got 55/25
you know p. Can you not figure its length?
√(33^2+44^2) = 55
To solve these problems, we'll need to understand some basic concepts in linear algebra. Let's go step by step:
1) To find the projection p of vector u onto vector w, we'll use the formula:
p = ((u dot w) / (w dot w)) * w
a) Calculate the dot product of u and w:
u dot w = (1 * 3) + (2 * 4) = 11
Calculate the dot product of w and w:
w dot w = (3 * 3) + (4 * 4) = 25
Now, use these values in the projection formula:
p = ((11) / (25)) * [3; 4]
p = [(11/25) * 3; (11/25) * 4]
p = [1.32; 1.76]
So, you were correct in finding p = [1.32; 1.76].
b) To find a scalar k such that the norm of the vector kp is equal to one, let's use the formula for the norm (also known as the Euclidean norm or the magnitude) of a vector:
||v|| = sqrt(v dot v)
Let's assume k is the scalar we are looking for. Therefore, we need to find k such that ||kp|| = 1.
Substitute v = kp into the formula:
||kp|| = sqrt((kp) dot (kp))
Using the properties of the dot product, we can rewrite this as:
||kp|| = sqrt(k^2 * (p dot p))
Now, equate this to 1:
1 = sqrt(k^2 * (p dot p))
Square both sides:
1^2 = (k^2 * (p dot p))
Simplify:
1 = k^2 * (p dot p)
Now, substitute the given value of p = [1.32; 1.76]:
1 = k^2 * (1.32^2 + 1.76^2)
Solve for k:
k^2 = 1 / (1.32^2 + 1.76^2)
k = sqrt(1 / (1.32^2 + 1.76^2))
Calculate the value:
k ≈ sqrt(1 / 5.68) ≈ 0.4462
So, the scalar k is approximately 0.4462, which will make the norm of the vector kp equal to one.
c) To find a vector that is orthogonal to the span of vector w, we need to find a vector s such that s is perpendicular to w.
A vector that is orthogonal to another vector can be found by taking the negative reciprocal of the components of the original vector. So, if w = [a; b], then s = [-b; a] would be orthogonal to w.
Given w = [3; 4], we can find s = [-4; 3].
Therefore, the vector s = [-4; 3] is orthogonal to the span {w}.
I hope this helps!