Question

Find the unit vector perpendicular to each of the vectors $3\widehat i + 2\widehat j + 3\widehat k$ and $3\widehat i - 2\widehat k$.

Answer 1

If we multiply two vectors by the methods of cross product (vector product) then we get the vector that is perpendicular to both parental vectors. That two parental vectors must not be equal to zero. Then divide that vector by the magnitude of the same vector in order to get a unit vector perpendicular to the two vectors.

Complete step-by-step solution:
Let, $\overrightarrow a = 3\widehat i + 2\widehat j + 3\widehat k$ and
$\overrightarrow b = 3\widehat i - 2\widehat k$
Find the cross product of $\overrightarrow a$ and $\overrightarrow b$
$\overrightarrow a \times \overrightarrow b$ is the determinant of the matrix \left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ 3&2&3 \\\ 3&0&{ - 2} \end{array}} \right)

{\widehat i}&{\widehat j}&{\widehat k} \\\ 3&2&3 \\\ 3&0&{ - 2} \end{array}} \right)$$ On taking determinant we get the result $$ = \widehat i( - 4 - 0) - \widehat j( - 6 - 9) + \widehat k(0 - 6)$$ On further solving $$ = - 4\widehat i + 15\widehat j - 6\widehat k$$ Let, $$\overrightarrow c = \overrightarrow a \times \overrightarrow b $$ So $$\overrightarrow c = - 4\widehat i + 15\widehat j - 6\widehat k$$ $$\overrightarrow c $$ is not a unit vector but we have to find a unit vector. So, to make this as a unit vector we have to divide by its magnitude. Magnitude of $$\overrightarrow c $$. $$|\overrightarrow c | = \sqrt {{{\left( { - 4} \right)}^2} + {{(15)}^2} + {{( - 6)}^2}} $$ On solving power $$|\overrightarrow c | = \sqrt {16 + 225 + 36} $$ $$|\overrightarrow c | = \sqrt {277} $$ Unit vector in the direction of $$\overrightarrow c $$ Unit vector $$\widehat c$$ in direction of $$\overrightarrow c $$ $$\widehat c = \dfrac{{\overrightarrow c }}{{|\overrightarrow c |}}$$ On putting the values $$\widehat c = \dfrac{{ - 4\widehat i + 15\widehat j - 6\widehat k}}{{\sqrt {277} }}$$ $$\widehat c = \dfrac{{ - 4}}{{\sqrt {277} }}\widehat i + \dfrac{{15}}{{\sqrt {277} }}\widehat j - \dfrac{6}{{\sqrt {277} }}\widehat k$$ Final answer: **The unit vector in the direction perpendicular to $$3\widehat i + 2\widehat j + 3\widehat k$$ and $$3\widehat i - 2\widehat k$$. $$ \Rightarrow \widehat c = \dfrac{{ - 4}}{{\sqrt {277} }}\widehat i + \dfrac{{15}}{{\sqrt {277} }}\widehat j - \dfrac{6}{{\sqrt {277} }}\widehat k$$** **Note:** Other than this type of multiplication, vectors are multiplied in many ways that are multiplication of a vector by a constant term. The dot product of two vectors that give the scalar quantity on multiplication. Next is the triple vector product and triple scalar product that will give scalar quantity and vector quantity respectively. In this particular question, you may make a mistake in finding the vector product of the two vectors. In the dot product, we have to multiply x with x, y with y, and z with z and add them to get the result of the dot product.