anonymous
  • anonymous
ques
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
ques
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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 \[p \vec x+q \vec y=\vec a\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.