anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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 ??
ganeshie8
  • ganeshie8
By any chance is it given that \(a\) and \(b\) are orthogonal ?
anonymous
  • anonymous
Nope, but there was another equation that u were given to prove before this one and I proved it it was \[p \vec x+q \vec y=\vec a\]

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ganeshie8
  • ganeshie8
Hmm that means \(a, x, y\) are coplanar
anonymous
  • 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
ganeshie8
  • ganeshie8
\(\vec x \times \vec a=q \vec b\) this does imply that \(a\perp b\) right ?
anonymous
  • anonymous
Ahh, yes!!!! \[\implies \vec b \perp (\vec a, \vec x)\]
anonymous
  • anonymous
then my first term reduces to 0
anonymous
  • anonymous
thanks :)
ganeshie8
  • ganeshie8
i believe so.. :)
ganeshie8
  • ganeshie8
np

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