Question

How do you find a unit vector perpendicular to two vectors that is perpendicular to both the vectors $u(0,2,1)$ and $v(1, - 1,1)$?

Answer 1

In cross product (or vector product) of two non-zero vectors $\overrightarrow u$ and $\overrightarrow v$, the resultant vector is perpendicular to both vectors$\overrightarrow u$ and $\overrightarrow v$. So here, first we need to find the cross product of two vectors. Remember that the resultant vector may or may not be a unit vector. Then, we need to find the magnitude and then the use the unit vector formula i.e. $\widehat w = \dfrac{{\overrightarrow w }}{{\left| {\overrightarrow w } \right|}}$. This means unit vectors are equal to vectors divided by magnitude of vectors.

Complete step by step answer:
Given two vectors$\overrightarrow u$ and $\overrightarrow v$. So, $\overrightarrow u \times \overrightarrow v$ is a vector that is perpendicular to both $\overrightarrow u$ and $\overrightarrow v$. Find the cross product of $u(0,2,1)$ and $v(1, - 1,1)$ i.e. $\widehat u = 2\widehat j + \widehat k$ and $\widehat v = \widehat i - \widehat j + \widehat k$. So, the determinant of the matrix will be:

{\widehat i}&{\widehat j}&{\widehat k} \\\ 0&2&1 \\\ 1&{ - 1}&1 \end{array}} \right|$$ Let $$\overrightarrow w = \overrightarrow u \times \overrightarrow v $$ $$\overrightarrow w = \widehat i(2 + 1) - \widehat j(0 - 1) + \widehat k(0 - 2)$$ $$ \Rightarrow \overrightarrow w = \widehat i(3) - \widehat j( - 1) + \widehat k( - 2)$$ $$ \Rightarrow \overrightarrow w = 3\widehat i + \widehat j - 2\widehat k$$ Next, we will find the magnitude of$$\overrightarrow w $$with its components, we use Pythagoras theorem which is written as $$\left| {\overrightarrow w } \right|$$. $$\left| {\overrightarrow w } \right|=\sqrt {{{(3)}^2} + {{(1)}^2} + {{( - 2)}^2}} $$ $$\Rightarrow \left| {\overrightarrow w } \right|= \sqrt {9 + 1 + 4} $$ $$\Rightarrow \left| {\overrightarrow w } \right|= \sqrt {14} $$ Now, we will find the unit vector of$$\widehat w$$. $$\widehat w = \dfrac{{\overrightarrow w }}{{\left| {\overrightarrow w } \right|}}$$ Substituting the values here, we get, $$\widehat w = \dfrac{{3\widehat i + \widehat j - 2\widehat k}}{{\sqrt {14} }} \\\ \Rightarrow \widehat w = \dfrac{3}{{\sqrt {14} }}\widehat i + \dfrac{1}{{\sqrt {14} }}\widehat j - \dfrac{2}{{\sqrt {14} }}\widehat k $$ This is a unit vector in component format. If we want to write the unit vector in bracket format, then it can be written as: $$\widehat w = (\dfrac{3}{{\sqrt {14} }},\dfrac{1}{{\sqrt {14} }},\dfrac{{ - 2}}{{\sqrt {14} }})$$ Hence, the unit vector perpendicular to$$u(0,2,1)$$and$$v(1, - 1,1)$$ both is$$\dfrac{3}{{\sqrt {14} }}\widehat i + \dfrac{1}{{\sqrt {14} }}\widehat j - \dfrac{2}{{\sqrt {14} }}\widehat k$$. We can check the final answer by finding the magnitude of the vector$$\widehat w$$. $$\left| {\widehat w} \right| = \sqrt {{{(\dfrac{3}{{\sqrt {14} }})}^2} + {{(\dfrac{1}{{\sqrt {14} }})}^2} + {{(\dfrac{{ - 2}}{{\sqrt {14} }})}^2}} $$ $$\Rightarrow \left| {\widehat w} \right| = \sqrt {\dfrac{9}{{14}} + \dfrac{1}{{14}} + \dfrac{4}{{14}}} $$ $$\Rightarrow \left| {\widehat w} \right|= \sqrt {\dfrac{{9 + 1 + 4}}{{14}}} \\\ \therefore \left| {\widehat w} \right|= \sqrt {\dfrac{{14}}{{14}}} $$ **Hence, a unit vector perpendicular to two vectors that is perpendicular to both the vectors $$u(0,2,1)$$ and $$v(1, - 1,1)$$ is $\sqrt {\dfrac{{14}}{{14}}}$.** **Note:** The unit vector$$\widehat i$$, $$\widehat j$$and$$\widehat k$$are along the direction of x-axis, y-axis and z-axis respectively. Because unit vectors are those whose magnitude is$$1$$. A vector quantity bears both magnitude and direction. It also takes the negative value when the object is in the opposite direction. Any two vectors must not be considered as equal just because they have equal magnitudes, because the direction in which the vectors are taken may vary. Thus, the unit vector specifies both magnitude and direction.