## anonymous one year ago ques

1. anonymous

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 ??

2. ganeshie8

By any chance is it given that $$a$$ and $$b$$ are orthogonal ?

3. anonymous

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

4. ganeshie8

Hmm that means $$a, x, y$$ are coplanar

5. anonymous

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

6. ganeshie8

$$\vec x \times \vec a=q \vec b$$ this does imply that $$a\perp b$$ right ?

7. anonymous

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

8. anonymous

then my first term reduces to 0

9. anonymous

thanks :)

10. ganeshie8

i believe so.. :)

11. ganeshie8

np

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