Question

Find a unit vector perpendicular to both the vectors $\overrightarrow a$ and $\overrightarrow b$, where $\overrightarrow a = \hat i - 7\hat j + 7\hat k$ and $\overrightarrow b = 3\hat i - 2\hat j + 2\hat k$?

Answer 1

We'll start by looking at the definition of vectors in order to answer the problem. The perpendicular vector will then be obtained by taking the cross – product of the vectors given in the question. After that, we'll divide the resulting vector by its magnitude to discover the unit vector.

Complete step by step answer:
Before we can answer this question, we must first understand what vectors are: A vector is a two-dimensional object or entity with a magnitude and a direction. In other words, a vector is a directed line segment with an arrow denoting the direction and a length equal to the magnitude of the vector.
We'll use the letters $\overrightarrow a$ and $\overrightarrow b$ to represent the first and second vectors, respectively.
We have given,
$\overrightarrow a = \hat i - 7\hat j + 7\hat k$
$\overrightarrow b = 3\hat i - 2\hat j + 2\hat k$
The vector that is perpendicular to both vectors a and b is obtained by cross-product of these vectors. The cross-product of two vector $\hat A = \dfrac{{(\hat j + \hat k)}}{{\sqrt 2 }}$ and $\vec B = x\hat i + y\hat j + z\hat k$ is given by

\hat i\,\,\,\,\,\,\hat j\,\,\,\hat k\,\,\, \\\ p\,\,\,\,q\,\,\,\,r \\\ x\,\,\,\,y\,\,\,\,z \\\ \end{gathered} \right|$$ Thus, the cross product of $$\overrightarrow a $$ and $$\overrightarrow b $$ will be given by $$\overrightarrow A \times \overrightarrow B = \left| \begin{gathered} \hat i\,\,\,\,\,\,\hat j\,\,\,\,\,\,\hat k\,\,\, \\\ 1\,\,\,\, - 7\,\,\,\,7 \\\ 3\,\,\,\, - 2\,\,\,\,2 \\\ \end{gathered} \right|$$ We have substituted the value and on solving we get, $$\overrightarrow a \times \overrightarrow b = \hat i[( - 7)(2) - (7)( - 2)] - \hat j[(1)(2) - (7)(3)] + \hat k[(1)( - 2) - ( - 7)(3)]$$ $$\overrightarrow a \times \overrightarrow b = \hat i[ - 14 + 14] - \hat j[2 - 21] + \hat k[ - 2 + 21]$$ $$\overrightarrow a \times \overrightarrow b = 19(\hat j + \hat k)$$ The cross product of two given vectors is $$19(\hat j + \hat k)$$. However, we are asked to find a perpendicular vector that is also a unit vector. We can get unit vectors by dividing a vector by its magnitude. Thus, the unit vector of the vector $$\vec A = p\hat i + q\hat j + r\hat k$$ is given by formula $$\hat A = \dfrac{{p\hat i + q\hat j + r\hat k}}{{\sqrt {{p^2} + {q^2} + {r^2}} }}$$ $$\hat A$$is the unit vector of $$\vec A$$ . Here we have p=0, q=19, k=19 $$\hat A = \dfrac{{0\hat i + 19\hat j + 19\hat k}}{{\sqrt {{0^2} + {{19}^2} + {{19}^2}} }}$$ $$\hat A = \dfrac{{19(\hat j + \hat k)}}{{19\sqrt 2 }}$$ On further solving, $$\hat A = \dfrac{{(\hat j + \hat k)}}{{\sqrt 2 }}$$ **The unit vector perpendicular to the $$\overrightarrow a = \hat i - 7\hat j + 7\hat k$$ and $\overrightarrow b = 3\hat i - 2\hat j + 2\hat k$ is $$\dfrac{{(\hat j + \hat k)}}{{\sqrt 2 }}$$ . **Note:** The cross product of two nonzero vectors is the product of the magnitude of both vectors and the sine of angle between the two vectors. We should be careful that the cross product of two vectors is the determinant of the matrix $$\left[ {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {{a_1}}&{{a_2}}&{{a_3}} \\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right]$$ of vectors, not their dot product which is $${a_1}{b_1} + {a_2}{b_2} + {c_1}{c_2}$$ .