Question

Given are non-zero non-orthogonal vectors \(\mathbf{A}, \mathbf{B}\). \(\mathbf{B}\) has components parallel and perpendicular to \(\mathbf{A}\) Prove using fundamental ideas and not simply by the manipulation of identities that \(\mathbf{B}\) can be written as:
\[\mathbf{B}=\frac{\mathbf{A}(\mathbf{A}\cdot\mathbf{B})}{|\mathbf{A}|^2}+\frac{(\mathbf{A}\times\mathbf{B})\times\mathbf{A}}{|\mathbf{A}|^2}\]
I'm having a little bit of trouble deriving the second part. Any hints?

Answer 1

\[ \frac{(A \times B )\times A}{|A|^2} = -\frac{(A \cdot B)A - (A \cdot A)B}{|A|^2}\]

Answer 2

We were told not to use that identity.

Answer 3

i guess i can't help it

Answer 4

What do you mean by fundamental ideas?

Answer 5

I mean things like the concept of vector addition, subtraction, dot product, cross product, scalar projections, vector projections. Also, the following were allowed:
\[\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\\
|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\]
But no other identities are.

Answer 6

Maybe this could lead somewhere fruitful - you are told that \(\mathbf{B}\) has some components parallel to \(\mathbf{A}\) and some which are perpendicular to it. So you could write \(\mathbf{B}\) as:\[\mathbf{B}=\alpha\mathbf{A}+\mathbf{C}\]where \(\mathbf{C}\) is perpendicular to \(\mathbf{A}\) such that:\[\mathbf{C}\cdot\mathbf{A}=0\]

Answer 7

BTW: \(\alpha\) is just some scalar constant.
For example, using this we get:\[\mathbf{A}\cdot\mathbf{B}=\mathbf{A}\cdot(\alpha\mathbf{A}+\mathbf{C})=\alpha\mathbf{A}\cdot\mathbf{A}+\mathbf{A}\cdot\mathbf{C}=\alpha|\mathbf{A}|^2\]

Answer 8

\[\therefore\frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}|^2}=\alpha\]

Answer 9

\[\therefore\frac{\mathbf{A}(\mathbf{A}\cdot\mathbf{B})}{|\mathbf{A}|^2}=\alpha\mathbf{A}\]

Answer 10

Now you just need to show that:\[\frac{(\mathbf{A}\times\mathbf{B})\times\mathbf{A}}{|\mathbf{A}|^2}=\mathbf{C}\]and you have proved the identity

Answer 11

And therein lies my problem.
I derived the first part a little differently than you did, but it gives the same result.

Answer 12

\[\mathbf{A}\times\mathbf{B}=\mathbf{A}\times(\alpha\mathbf{A}+\mathbf{C})=\alpha\mathbf{A}\times\mathbf{A}+\mathbf{A}\times\mathbf{C}=|\mathbf{A}||\mathbf{C}|\mathbf{y}\]where \(\mathbf{y}\) isa unit vector perpendicular to \(\mathbf{A}\) and \(\mathbf{C}\).
Therefore:\[(\mathbf{A}\times\mathbf{B})\times\mathbf{A}=|\mathbf{A}||\mathbf{C}|\mathbf{y}\times\mathbf{A}=|\mathbf{A}||\mathbf{C}||\mathbf{y}||\mathbf{A}|\mathbf{c}=|\mathbf{A}|^2|\mathbf{C}|\mathbf{c}=|\mathbf{A}|^2\mathbf{C}\]therefore:\[\frac{(\mathbf{A}\times\mathbf{B})\times\mathbf{A}}{|\mathbf{A}|^2}=\mathbf{C}\]

Answer 13

where \(\mathbf{c}\) is the unit vector in the direction of \(\mathbf{C}\)

Answer 14

Wait, I don't understand why
\[\mathbf{A}\times\mathbf{C}=|\mathbf{A}||\mathbf{C}|\mathbf{y}\]

Answer 15

its because \(\mathbf{A}\) and \(\mathbf{C}\) are perpendicular to one another

Answer 16

\[|\mathbf{A}\times\mathbf{C}|=|\mathbf{A}||\mathbf{C}|\sin(90)=|\mathbf{A}||\mathbf{C}|\]

Answer 17

Ah, so essentially, the trick is to express \(\mathbf{B}=\alpha\mathbf{A}+\mathbf{C}\) from the get-go. Thanks for the help!

Answer 18

yw :)