Given two vectors $\vec{a} = (2, 3), \vec{b} = (4, -1)$. Project vector B onto vector A, get the new vector's coordinates and its length.

I managed to find the length first but it didn't quite match up with the answer with the one in the book. How do you do these kind of questions?

Question

Given two vectors $\vec{a} = (2, 3), \vec{b} = (4, -1)$. Project vector B onto vector A, get the new vector's coordinates and its length.

I managed to find the length first but it didn't quite match up with the answer with the one in the book. How do you do these kind of questions?

ganeshie8 · Answer

projection of $\vec{b}$ along $\vec{a}$ =    $\large  \vec{b} . \hat{a}$

tyteen4a03 · Answer

@ganeshie8 That was surprisingly easy - no idea why the resources I found overcomplicates it. Thanks!

ganeshie8 · Answer

:) yup its easy once u see it !

tyteen4a03 · Answer

@ganeshie8 But that's the scalar product though. What would $\vec{b_a}$ be then?

ganeshie8 · Answer

multiply the magnitude $\vec{b} . \hat{a}$ wid the unit vector $\hat{a}$

ganeshie8 · Answer

$(\vec{b} . \hat{a} ) \hat{a}$

ganeshie8 · Answer

that doesnt make sense is it

tyteen4a03 · Answer

$\hat{a} = (-y, x)$ according to my useless textbook. It apparently rotates the vector by 90 degrees clockwise.

ganeshie8 · Answer

|dw:1384373120657:dw|

ganeshie8 · Answer

suppose those are the two given vectors $\vec{a}$ and $\vec{b}$

ganeshie8 · Answer

|dw:1384373207994:dw|

ganeshie8 · Answer

clearly vector projection of a along b =  $(a\cos \theta )\hat{b} = (\vec{a} . \hat{b})\hat{b} $

ganeshie8 · Answer

ur textbook thing makes no sense to me lol

tyteen4a03 · Answer

@ganeshie8 What should \hat{a} stand for? Unit vector?

ganeshie8 · Answer

Yes ! $\hat{a}$ is the unit vector along $\vec{a}$

ganeshie8 · Answer

$\hat{b}$ is the unit vector along $\vec{b}$

ganeshie8 · Answer

projection formula above follows from the fact that  mag of unit vector is 1....

ganeshie8 · Answer

lets work the present problem may be

ganeshie8 · Answer

$\vec{a} = (2, 3), \vec{b} = (4, -1) $

tyteen4a03 · Answer

$$\vec{e_a}$$
		$$\begin{Bmatrix}
		\frac{2}{\sqrt{2^2 + 3^2}} \ \frac{3}{\sqrt{2^2 + 3^2}}
		\end{Bmatrix}$$
		$$=\begin{Bmatrix}
		\frac{2}{\sqrt{13}} \ \frac{3}{\sqrt{13}}
		\end{Bmatrix}$$

		$$|\vec{b_a}| = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}|}$$
		$$=\frac{|(2)(4) - 3|}{\sqrt{4^2 + (-1)^2}}$$
		$$=\frac{5}{\sqrt{17}}$$

This is what I have. e_a = unit vector of a.

ganeshie8 · Answer

you made a mistake at 2nd step from last

ganeshie8 · Answer

bottom u shud take mag of a
instead, u took mag of b

ganeshie8 · Answer

$\large  |\vec{b_a}| = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}|}$

$\large  |\vec{b_a}| = \frac{|(2)(4)-3|}{\sqrt{2^2+3^2}}$

tyteen4a03 · Answer

That was a silly mistake... Thanks :P

ganeshie8 · Answer

np :)