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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is given that \[\vec x =\lambda \vec a+\frac{q(\vec a \times \vec b)}{a^2}\] and \[\vec y =\frac{(1p \lambda)}{q}\vec a\frac{p(\vec a \times \vec b)}{a^2}\] 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 \[(\lambda \vec a+\frac{q(\vec a \times \vec b)}{a^2}) \times(\frac{(1p \lambda)\vec a}{q}\frac{p(\vec a \times \vec b)}{a^2})\] \[\frac{\lambda(1p \lambda)}{q}(\vec a \times \vec a)+\frac{(1 p \lambda)}{a^2}(\vec c \times \vec a)\frac{p \lambda}{a^2}(\vec a \times \vec c)\frac{pq}{a^4}(\vec c \times \vec c)\] where I've taken \[\vec c =\vec a \times \vec b\] The first and last term becomes 0 \[\frac{(p \lambda1)}{a^2}(\vec a \times \vec c)\frac{p \lambda}{a^2}(\vec a \times \vec c)=(\vec a \times \vec c)(\frac{p \lambda1p \lambda}{a^2})\] \[(\vec a \times \vec c)(\frac{1}{a^2})=(\vec a \times (\vec a \times \vec b))(\frac{1}{a^2})\]\[[(\vec a . \vec b)\vec a(\vec a . \vec a)\vec b](\frac{1}{a^2})\] After expanding we will get the 2nd term as vector b but what about the first term?? Where am I making a mistake ??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2By any chance is it given that \(a\) and \(b\) are orthogonal ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope, but there was another equation that u were given to prove before this one and I proved it it was \[p \vec x+q \vec y=\vec a\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Hmm that means \(a, x, y\) are coplanar

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0They 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
 one year ago
Best ResponseYou've already chosen the best response.2\(\vec x \times \vec a=q \vec b\) this does imply that \(a\perp b\) right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, yes!!!! \[\implies \vec b \perp (\vec a, \vec x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then my first term reduces to 0
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