# Minimum and maximum values of $(\vec{r}\times\hat{j}).\vec{c}$

If unit vector $\vec{c}$ makes an angle $\frac{\pi}{3}$ with $\hat{i}+\hat{j}$ ,then minimum and maximum values of $(\vec{r}\times\hat{j}).\vec{c}$ respectively are

1. $0,\frac{\sqrt3}{2}$
2. $\frac{-1}{2},\frac{\sqrt3}{2}$
3. $\frac{-\sqrt3}{2},\frac{\sqrt3}{2}$
4. none of these

$$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \\ (\vec{r}\times\hat{j}).\vec{c} =(x\hat{i}+y\hat{j}+z\hat{k})\times(\hat{j}).\vec{c} \\ =(x\hat{k}-z\hat{i}).\vec{c}$$

But i am stuck here.

• HINT: What would be $\hat{k} \cdot \vec{c}$? – hjpotter92 Nov 9 '15 at 9:48
• Since there is no constraint on $\vec{r}$ there wont be a maximum or minimum – G-man Nov 9 '15 at 9:52

The constraint on $\vec c$ means that its tip lies on a circle lying in a plane with normal $\vec i + \vec j$.
The quantity $(\vec{r}\times\hat{j}).\vec{c}$ is the (signed) volume of the parallelipiped with sides $\vec r$, $\vec j$, $\vec c$. By changing $\vec r$, you can obtain any number you want for this volume.
• If we let $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$,then according to the given constraint on $\vec{c}$,$c_1+c_2=\frac{1}{\sqrt2}$,we need to find the max and min value of $\vec{r}\times\hat{j}.\vec{c}=c_3x-c_1z$.How do we find the max and min value of $c_3x-c_1z$ under constraints $c_1+c_2=\frac{1}{\sqrt2}$ and $c_1^2+c_2^2+c_3^2=1?$@bubba – diya Jan 26 '16 at 2:54