Question

ques

Answer 1

It is given that
\[\vec x =\lambda \vec a+\frac{q(\vec a \times \vec b)}{a^2}\]
and
\[\vec y =\frac{(1-p \lambda)}{q}\vec a-\frac{p(\vec a \times \vec b)}{a^2}\]
where
\[a=|\vec a|\]
and lambda is any scalar
We have to prove that vector x and y satisfy the equation :
\[\vec x \times \vec y=\vec b\]
Now here's what I done but I just can't find my mistake
\[(\lambda \vec a+\frac{q(\vec a \times \vec b)}{a^2}) \times(\frac{(1-p \lambda)\vec a}{q}-\frac{p(\vec a \times \vec b)}{a^2})\]
\[\frac{\lambda(1-p \lambda)}{q}(\vec a \times \vec a)+\frac{(1- p \lambda)}{a^2}(\vec c \times \vec a)-\frac{p \lambda}{a^2}(\vec a \times \vec c)-\frac{pq}{a^4}(\vec c \times \vec c)\]
where I've taken
\[\vec c =\vec a \times \vec b\]
The first and last term becomes 0
\[\frac{(p \lambda-1)}{a^2}(\vec a \times \vec c)-\frac{p \lambda}{a^2}(\vec a \times \vec c)=(\vec a \times \vec c)(\frac{p \lambda-1-p \lambda}{a^2})\]
\[(\vec a \times \vec c)(\frac{-1}{a^2})=(\vec a \times (\vec a \times \vec b))(\frac{-1}{a^2})\]\[[(\vec a . \vec b)\vec a-(\vec a . \vec a)\vec b](\frac{-1}{a^2})\]
After expanding we will get the 2nd term as vector b but what about the first term??
Where am I making a mistake ??

Answer 2

By any chance is it given that \(a\) and \(b\) are orthogonal ?

Answer 3

Nope,
but there was another equation that u were given to prove before this one and I proved it
it was
\[p \vec x+q \vec y=\vec a\]

Answer 4

Hmm that means \(a, x, y\) are coplanar

Answer 5

They have done it like this in the solution
first they crossed equation 1 with x
\[\vec x \times (p \vec x+q \vec y)=\vec x \times \vec a\]\[q(\vec x \times \vec y)=\vec x \times \vec a\]\[\vec x \times \vec a=q \vec b\]
Now they crossed with a
\[\vec a \times (\vec x \times \vec a)=\vec a \times q \vec b\]\[(\vec a . \vec a)\vec x-(\vec a . \vec x)\vec a=q(\vec a \times \vec b)\]\[a^2 \vec x=(\vec a . \vec x)\vec a+q(\vec a \times \vec b)\]\[\vec x=\frac{(\vec a . \vec x) \vec a}{a^2}+\frac{q(\vec a \times \vec b)}{a^2}\]\[\vec x =\lambda \vec a+\frac{q(\vec a \times \vec b)}{a^2}\]
they substituted this back in equation 1 and solved for y, which was coming correct

Answer 6

\(\vec x \times \vec a=q \vec b\)

this does imply that \(a\perp b\) right ?

Answer 7

Ahh, yes!!!! \[\implies \vec b \perp (\vec a, \vec x)\]

Answer 8

then my first term reduces to 0

Answer 9

thanks :)

Answer 10

i believe so.. :)

Answer 11

np