Consider three vectors $\vec{p},\vec{q},\vec{r}$ such that
$\vec{p}\times\vec{r}=\vec{q}+c\vec{p}$ and $\vec{p}.\vec{r}=2$.Find the vector $\vec{r}$ and the value of $c$.

My Attempt:
As $\vec{p}\times\vec{r}=\vec{q}+c\vec{p}$,so $\vec{r}$ must be in the perpendicular to both $\vec{p}$ and $\vec{q}$.So let $\vec{r}=\lambda( \vec{p}\times\vec{q})$
I calculated $\vec{r}=\lambda(2\hat{i}-2\hat{k})=2\lambda\hat{i}-2\lambda\hat{k}$
But $\vec{p}.\vec{r}$ comes out to be $0$ and not $2$,as given in one of the conditions.What is wrong in this approach?

My Second try:
As $\vec{p}$ and $\vec{q}$ are non-collinear vectors,So
Let $\vec{r}=x\vec{p}+y\vec{q}+z\vec{p}\times\vec{q}$

But i could not solve it further.Please help me.

  • $\begingroup$ Try $(\vec{p} \times \vec{r}). \vec{r}$ and $(\vec{p} \times \vec{r}). \vec{p}$ and see the results . $\endgroup$ – Nizar Nov 16 '15 at 16:22

Looks pretty straightforward to me: $$ \vec p=(1,1,1) , \vec q=(1,-1,1)$$ $$\vec p \times \vec r=\left[\begin{array}{ccc}1&1&1\\ r_{a}&r_{b}&r_{c} \end{array} \right]=(r_{c}-r_{b})\hat i+(r_{a}-r_{c})\hat j+(r_{b}-r_{a})\hat k $$ $$\vec q+c\vec p=(1+c,-1+c,1+c)$$ $$r_{c}-r_{b}=1+c ,r_{a}-r_{c}=-1+c,r_{b}-r_{a}=1+c$$ $$\vec r.\vec p= r_{a}+ r_{b}+r_{c}=2$$

Then you solve the above system which gives you $$\vec r=(0,2/3,4/3), c=-1/3$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.