Question

Given \(\large \color{green}{ A = \frac{10}{\sqrt{2}}(a_x + a_z)}\) and \( \large \color{blue}{B = 3(a_y + a_z)}\), express the projection of \(B\) on \(A\) as a vector in the direction of \(A\)..

Answer 1

Isn't it something like
\[\text{proj}_AB=\frac{A\cdot B}{|A|}~~?\]

Answer 2

Great, I thought I'd forgotten it :)

Answer 3

But that will give you a scalar quantity at then end I think..

Answer 4

yes

Answer 5

IT is scalar product

Answer 6

But Final answer is a vector in the book..

Answer 7

@ganeshie8

Answer 8

@Kainui

Answer 9

Sorry, but that was meaningless....

Answer 10

Oh sorry, I think it's multiplied by \(A\). That makes a scalar multiple of the vector \(A\), and so it's in the same direction.
\[\frac{A\cdot B}{|A|}A\]

Answer 11

Hey, don't tag anybody, that is my work.. :P

Answer 12

Yeah i know xD, I started vectors just a few days back

Answer 13

Ganesh has seen it, may be he is trying on his first..

Answer 14

You have multiplied vector A there???

Answer 15

Okay I try that first..

A.B = 30/sqrt{2}

Answer 16

if you want vector along A, then multiply the mag by unit vector of A ?

Answer 17

|A| = 10

Dividing we get 3/sqrt{2}

Answer 18

And then multiplying by again vector A will give me:

30/sqrt{2}(a_x + a_z) and this is not the answer...

Answer 19

Yeah, so you should have
\[A\cdot B=\frac{30}{\sqrt2}a_xa_y+\frac{30}{\sqrt2}{a_z}^2\]
\[|A|=\sqrt{\left(\frac{30}{\sqrt2}a_x\right)^2+\left(\frac{30}{\sqrt2}a_z\right)^2}=\frac{30}{\sqrt2}\sqrt{{a_x}^2+{a_z}^2}\]
\[\frac{A\cdot B}{|A|}=\frac{a_xa_y+{a_z}^2}{\sqrt{{a_x}^2+{a_z}^2}}\]
\[\frac{A\cdot B}{|A|}A={\large\frac{30\left(a_xa_y+{a_z}^2\right)}{\sqrt2\sqrt{{a_x}^2+{a_z}^2}}(a_x+a_z)}\]

Answer 20

Wait Wait Wait..

How you got |A| that complex??

Answer 21

I used the Pythagorean theorem. Isn't \(A\) given with variable coordinates?

Answer 22

Oh sorry, May be you all did not get and I am sorry, I have not told you all earlier..

a_x, a_y and a_z are unit vectors along x, y and z direction..

Answer 23

Oh well in that case just replace all \(a_x,a_y,a_z\) with 1.

Answer 24

In A.B, a_x and a_y will become 0..

Answer 25

\[A.B = (\frac{10}{\sqrt{2}} \cdot 3)a_z.a_z\]

Answer 26

You want B's magnitude in A's direction.

So you divide out A's magnitude and find B's magnitude, and then multiply those together.

Answer 27

\[A.B = \frac{30}{\sqrt{2}}\]

Answer 28

Divide out A's magnitude with what??? @Kainui

Answer 29

With A's magnitude. \[\Large \frac{ \bar A}{| \bar A|}\] See, you just find the length of A, divide the vector A by the length and you will be left with a unit vector in the direction of A. It's essentially a directional form of the number 1 that you can multiply any scalar by to have this direction.

Answer 30

Sorry, guys I will come somewhat later, my dinner is waiting for me..

Then I try it..

And @Kainui By that yes I will get unit vector along A's direction, then you want me to multiply it with B's magnitude???

Answer 31

Okay, start over:
\[A=\frac{10}{\sqrt2}(a_x+a_z)=\frac{10}{\sqrt2}\bigg(\langle 1,0,0\rangle+\langle0,0,1\rangle\bigg)=\frac{10}{\sqrt 2}\langle 1,0,1\rangle\\
B=3(a_y+a_z)=3\bigg(\langle 0,1,0\rangle+\langle0,0,1\rangle\bigg)=3\langle0,1,1\rangle\]
Now,
\[A\cdot B=0+0+\frac{30}{\sqrt2}=\frac{30}{\sqrt 2}\\
|A|=\sqrt{\left(\frac{10}{\sqrt2}\right)^2+0^2+\left(\frac{10}{\sqrt2}\right)^2}=\sqrt{\frac{200}{2}}=10\]
\[\frac{A\cdot B}{|A|}A=\frac{3}{\sqrt2}A=\frac{30}{4}\langle1,0,1\rangle \]

Answer 32

I am afraid, that is not matching with my final answer.. @SithsAndGiggles

Answer 33

@waterineyes exactly, that's all that's left to do. That's actually what the formulas are saying, it's just like condensed down and tends to leave out reasoning

It's like we can think in terms of "square root of four" but the formula just says "2".

Answer 34

That 4 should be a 2... So you can reduce to
\[15\langle1,0,1\rangle\]

Answer 35

Sorry @SithsAndGiggles I didn't mean to be rude and come in.

Answer 36

No worries! It's nice to have more people around to tell me if I made a mistake :)

Answer 37

No, this way also I am not getting, @Kainui just solve it on your own and copy with my final answer, I am writing here...

Answer 38

\[\frac{A}{|A|} = \frac{(a_x + a_z)}{\sqrt{2}}\]

Answer 39

Correct.

Answer 40

\[\frac{A}{|A|} \cdot |B| = 3(a_x + a_z)\]

Answer 41

Yes. This is correct.

Answer 42

WAIT sorry.

Answer 43

Why??

Answer 44

This is not correct. We are not looking for a vector of length B in the direction of A. X_X

Answer 45

See the difference? The projection is the smaller one, the vector in the direction of A is much longer.

Answer 46

Wait if I do:

\[(B.a_A) \cdot a_A\]

Answer 47

This is still in the same direction as A, so A/|A| is still useful to us. But we can't multiply this by the magnitude of B.

Instead the length projection is the dot product of B and A.

Answer 48

We should again multiply it with the direction of A and direction of A is given by vector by its length..

Answer 49

Yes, the formula which Sith gave earlier is good, it just needs to be multiplied by the direction. Hahaha this is becoming an ordeal!

Answer 50

In your terms If I do:

\[(\frac{B. A}{|A|}) \cdot \frac{A}{|A|}\]

I think I am getting answer by this way..

Answer 51

Yes this looks good now haha.

Answer 52

Sith was saying to multiply with whole vector(Mag + Direc) but we have to multiply only with the direction...