Question

Find cosine of the angle between the direction and of the vectors $\vec a + 4\hat i + 3\hat j + \hat k,\,\,\,\vec b = 2\hat i - \hat j + 2\hat k$.
Also, find a unit vector perpendicular to both $\vec a\,\,and\,\,\vec b$. What is the sine of the angle between $\vec a\,\,and\,\,\vec b$.
A. $\sin \theta - \sqrt {\left( {\dfrac{{185}}{{26 \times 9}}} \right)} - \dfrac{1}{3}\sqrt {\left( {\dfrac{{185}}{{26}}} \right)}$
B. $\sin \theta = - \sqrt {\left( {\dfrac{{185}}{{26 \times 9}}} \right)} = \dfrac{1}{3}\sqrt {\left( {\dfrac{{185}}{{26}}} \right)}$
C. $\sin \theta - \sqrt {\left( {\dfrac{{185}}{{26 \times 8}}} \right)}$
D. None of these

Answer 1

To find cosine angle between two vectors, for sine angle between $\vec a\,\,and\,\,\vec b$, we use cross product of vectors. Also for unit vector, we divide vector by its magnitude as given below:
cos\theta = \dfrac{{\vec a\,.\vec b}}{{|\vec a\,|\,|\vec b|}},\,\,\,\sin \theta = \dfrac{{\vec a\, \times \vec b}}{{|\vec a\,|\,\,|\vec b|}}$$$$\vec a\,.\vec b = ({a_x}\hat i + {a_y}\hat j + {a_z}\hat k).({b_x}\hat i + {b_y}\hat j + {b_z}\hat k) = {a_x}{b_x} + {a_y}\,{b_y} + {a_z}{b_z}

{\hat i}&{\hat j}&{\hat k} \\\ {{a_x}}&{{a_y}}&{{a_z}} \\\ {{b_x}}&{{b_y}}&{{b_z}} \end{array}} \right| = \hat i({b_z}{a_y} - {a_z}{b_y}) - \hat j({a_x}{b_z} - {b_x}{a_z}) + \hat k({a_x}{b_y} - {b_x}{a_y})$$ **Complete step by step answer:** (1) Given vectors are $$\vec a\, = 4\hat i + 3\hat j + k,\,\,\vec b = 2\hat i - \hat j + 2\hat k$$ (2) For cosine of the angle, we use dot product $$\vec a\,\,.\,\vec b = \,|\vec a|\,\,|\vec b|\cos \theta $$ (3) $$\cos \theta = \dfrac{{\vec a.\vec b}}{{|\vec a|\,|\vec b|}}$$ (4) Calculate dot product of vector $$\vec a.\vec b = (4\hat i + 3\hat j + \hat k).(2\hat i - \hat j + 2\hat k)$$ Multiplying $$4\hat i \times 2\hat i$$,$$3\hat j \times - \hat j\,and\,\,\hat k \times 2\hat k$$ ,we get $$ = 8 - 3 + 2$$ $$ = 10 - 3$$ $$\vec a.\vec b = 7$$ Also,$$|\vec a|\,\, = \sqrt {{{(4)}^2} + {{(3)}^2} + {{(1)}^2}} $$ $ = \sqrt {16 + 9 + 1} $ $$ = \sqrt {26} $$ $$|\vec b|\,\, = \sqrt {{{(2)}^2} + {{(1)}^2} + {{(2)}^2}} $$ $$ = \sqrt {4 + 1 + 4} $$

= \sqrt 9 \\
= 3 \\

$$\left( 5 \right)$$ Using values of step $$\left( 4 \right)$$in step$$\left( 3 \right)$$, we get $$cos\theta = \dfrac{7}{{\sqrt {26} .(3)}}$$ Which is the required cosine of the angle between vectors. (6) Now to find unit vector perpendicular to both a and b is given as $$\dfrac{{(a \times b)}}{{|a \times b|}}$$ (7) Calculating $$\vec a \times \vec b$$first $$\vec a \times \vec b\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 4&3&1 \\\ 2&{ - 1}&2 \end{array}} \right|$$ $$ = \hat i(6 + 1) - \hat j(8 - 2) + \hat k( - 4 - 6)$$

|\vec a \times \vec b|, = 7\hat i - 6\hat j - 10k \\
|\vec a \times \vec b|, = \sqrt {{{(7)}^2} + {{( - 6)}^2} + {{( - 10)}^2}} \\

$$\sqrt {49 + 36 + 100} = \sqrt {285} $$ (8) Therefore, unit perpendicular vector to both $$\vec a\,\,and\,\,\vec b$$ is $$\dfrac{{7\hat i + 6\hat j - 10\hat k}}{{\left( {\sqrt {285} } \right)}}$$ (9) Now to calculate sinθ of the angle between the given vectors $$\vec a \times \vec b = \,|\vec a|\,\,\vec b|\,\sin \theta $$ $$ \Rightarrow \sin \theta = \dfrac{{\vec a \times \vec b}}{{|\vec a|\,\,|\vec b|}}$$ $$\sin \theta = \dfrac{{(7\hat i + 6\hat j - 10\hat k)}}{{\sqrt {26} \,\,\sqrt 3 }}$$ $$\sin \theta = \dfrac{{7\hat i + 6\hat j - 10\hat k}}{{\sqrt {78} }}$$ Which is required sinθ of the angle between given vectors. **Hence, the correct option is D.** **Note:** A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction.