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If there are 2 vectors A & B in 2D space and I want to project B onto A, My approach is to find the unit vector in the direction of A and multiply it with a scalar. Since the Dot product gives the amount of B in direction of A, the required scalar is given by the dot product between A & B.

Now, we have both direction and magnitude (scalar) for the projected vector.

But this is wrong compared to answers on the internet. I have to divide my result by the length of A again (i have done it once to find the unit vector in the direction of A). I do not understand why should I divide it again. Can someone please help me?

PS: Please don't use cosine to explain. I am trying to understand this as a topic of linear algebra which doesn't use trigonometric identities.

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  • $\begingroup$ You say the required scalar is given by "the dot product between $A$ and $B$." Where in this did you already divide by the length of $A$? Also: please show your exact work so we can see if something was done wrong. $\endgroup$
    – MPW
    Commented Jan 30, 2023 at 18:58
  • $\begingroup$ If you write $B=\lambda A+A_\perp$ with $\langle A,\,A_\perp\rangle=0$, you can show $\lambda=\frac{\langle A,\,B\rangle}{\Vert A\Vert^2}$. Is that ${}^2$ what you're asking about? $\endgroup$
    – J.G.
    Commented Jan 30, 2023 at 21:54
  • $\begingroup$ Yes, Can you please prove it? What is geometrical meaning of dividing the scalar quantity of dot product by the square of the length of the vector A? $\endgroup$
    – Psetty
    Commented Jan 31, 2023 at 0:24
  • $\begingroup$ If $\langle \hat{A},B\rangle$ is the scalar projection, then $\langle\hat{A},B\rangle\hat{A}$ is the vector projection. Since there are two $\hat{A}$'s and $\hat{A}=A/\|A\|$, this is equivalent to $\langle A,B\rangle A/\|A\|^2$. Note the projection of $B$ onto $A$ only depends on $A$'s direction, not its magnitude. If you had two $A$'s in the numerator but only one $\|A\|$ in the denominator, the resulting quantity would scale with $A$'s magnitude, which we don't want. $\endgroup$
    – coiso
    Commented Aug 28, 2023 at 16:06

1 Answer 1

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We want the projection of $B$ along $A$ and along $cA$ (for any $c \ne 0$) to be equal. Since $\langle{cA, B}\rangle = c \langle{A, B}\rangle$, we have that $\langle{cA, B}\rangle \cdot (cA) = c^2 \langle{A, B}\rangle\cdot A$. This suggests that we should divide by something proportional to $c^2$. The correct quantity happens to be $c^2||A||^2 = ||cA||^2 = \langle{cA, cA}\rangle$, as can be verified by taking $B = A$, in which case we want the result to be $A$. In summary, the required projection is $$\frac{\langle{cA, B}\rangle}{\langle{cA, cA}\rangle} cA = \frac{\langle{A, B}\rangle}{\langle{A, A}\rangle} A. $$

Alternatively, following the suggestion by @J.B., if $B = \lambda A + A^\perp$ with $\langle{A, A^\perp}\rangle = 0$ then $$\langle{A, B}\rangle = \langle{A, \lambda A}\rangle + \langle{A, A^\perp\rangle} = \lambda \langle A, A\rangle$$ so $\lambda = \frac{\langle{A, B}\rangle}{\langle{A, A}\rangle}$.

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