(a).
If v ∈ ℒ(w), that means v = kw for some constant w. In that case:
proj(w, v) = (v·w/(w·w))w = k(w·w/(w·w))w = kw = v
Thus, ℒ(v, w) = ℒ(v, proj(w, v)) = ℒ(w).
(b).
We have:
perp(w, v) = v - proj(w, v)
So:
ℒ(v, w) = ℒ(perp(w, v) + proj(w, v), w)
Since proj(w, v) and w are collinear:
ℒ(perp(w, v) + proj(w, v), w) = ℒ(perp(w, v), w)
We don't even need v to be outside of ℒ(w), this will always work. When v ∈ ℒ(w), that will further simplify to just ℒ(w).
#Please solve this...
wicked talon
glad cliff
(a).
If v ∈ ℒ(w), that means v = kw for some constant w. In that case:
proj(w, v...
So both a) and b) are true?
wicked talon
Yes, both seem to be true.
glad cliff
Thank you!
wicked talon
You're welcome!
glad cliff
Yes, both seem to be true.
By L(w) are you referring L as Span?
wicked talon
By L(w) are you referring L as Span?
Yeah, I usually denote span as ℒ (script L: stands for "linear span").
glad cliff
Yeah, I usually denote span as ℒ (script L: stands for "linear span").
Ah ok, thanks boss!
wicked talon
You're welcome!
ionic jackalBOT
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wicked talon
(c).
If s is a direction vector for the line, then ks for any nonzero constant k is also a direction.
But since we need a unit vector, there can only be two, just with different signs.
(d).
Two vectors have the same direction if their unit vector is the same (up to sign).
glad cliff
(c).
If s is a direction vector for the line, then ks for any nonzero constant k...
Oh okay. Ty. How would I do b)?
ionic jackalBOT
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glad cliff
(a).
If v ∈ ℒ(w), that means v = kw for some constant w. In that case:
proj(w, v...
How is your line, after you said In that case: true?