Eric Nicholas K's answer to Priya's Junior College 1 H2 Maths Singapore question.
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Maybe this?
For the geometrical proof, it’s probably something like this.
Projection vector of u onto v simply means a vector, in the direction of v, which has a certain length to be determined.
(Though in this case, v is not a unit vector in the direction of v - the projection formula vector specifically requires v-cap as part of the formula - but the arguments below still hold anyway; we will eventually prove that |v| is 1 later on somehow)
In this particular case, the projection vector of u onto v (which is supposed to be a vector in the direction of v)…
…happens to be u itself. This goes to show that u and v must be parallel vectors themselves.
For the geometrical proof, it’s probably something like this.
Projection vector of u onto v simply means a vector, in the direction of v, which has a certain length to be determined.
(Though in this case, v is not a unit vector in the direction of v - the projection formula vector specifically requires v-cap as part of the formula - but the arguments below still hold anyway; we will eventually prove that |v| is 1 later on somehow)
In this particular case, the projection vector of u onto v (which is supposed to be a vector in the direction of v)…
…happens to be u itself. This goes to show that u and v must be parallel vectors themselves.
Date Posted:
2 years ago
Sorry, ignore the part after "|v| = 1".
(i.e. ignore the last part from "since |v| = 1"...)
Those are irrelevant to the working required.
(i.e. ignore the last part from "since |v| = 1"...)
Those are irrelevant to the working required.
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