Question

Determine unit vector perpendicular to both a vector avector=2i+j+k and bvector =i+j+2k

Answer 1

When we take the cross product of two vectors, the resulting vector is perpendicular to both the vectors.


This is what we want to do. We have two vectors, and we want to find a unit vector perpendicular to these two vectors.


Let:


$\vec{a} = 2\hat{\imath}+\hat{\jmath}+\hat{k}\\ \\ \vec{b} = \hat{\imath}+\hat{\jmath}+2\hat{k}$


Let the cross product of these two vectors be $\vec{r}$

We have:


$\vec{r}=\vec{a}\times\vec{b}\\\\\\\implies\vec{r}=(2\hat{\imath}+\hat{\jmath}+\hat{k})\times(\hat{\imath}+\hat{\jmath}+2\hat{k})\\\\\\\implies\vec{r}=\left|\begin{array}{ccc}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&1&1\\1&1&2\end{array}\right|\\\\\\\implies\vec{r}=\hat{\imath}((1)(2)-(1)(1))-\hat{\jmath}((2)(2)-(1)(1))+\hat{k}((2)(1)-(1)(1))\\\\\\\implies\vec{r}=\hat{\imath}(2-1)-\hat{\jmath}(4-1)+\hat{k}(2-1)\\\\\\\implies\vec{r}=\hat{\imath}-3\hat{\jmath}+\hat{k}$



Now, we just need to find the unit vector for $\vec{r}$


To find the unit vector, we divide the vector by its magnitude.


$\vec{r}=\hat{\imath}-3\hat{\jmath}+\hat{k}\\\\\\\implies |\,\vec{r}\,|=\sqrt{1^2+(-3)^2+1^2}\\\\\\\implies |\,\vec{r}\,|=\sqrt{11}$


Now, we can easily find the unit vector:


$\displaystyle\vec{r}=\hat{\imath}-3\hat{\jmath}+\hat{k}\\\\\\\implies \hat{r}=\frac{\vec{r}}{|\,\vec{r}\,|}\\\\\\\implies\hat{r}=\frac{\hat{\imath}-3\hat{\jmath}+\hat{k}}{\sqrt{11}}\\\\\\\implies\boxed{\boxed{\boxed{\hat{r}=\frac{1}{\sqrt{11}}\,\hat{\imath}-\frac{3}{\sqrt{11}}\,\hat{\jmath}+\frac{1}{\sqrt{11}}\,\hat{k}}}}$



This is the unit vector perpendicular to both the given vectors.