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

Hint: 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:
\[cos\theta = \dfrac{{\vec a\,.\vec b}}{{|\vec a\,|\,|\vec b|}},\,\,\,\sin \theta = \dfrac{{\vec a\, \times \vec b}}{{|\vec a\,|\,\,|\vec b|}}\]\[\vec a\,.\vec b = ({a_x}\hat i + {a_y}\hat j + {a_z}\hat k).({b_x}\hat i + {b_y}\hat j + {b_z}\hat k) = {a_x}{b_x} + {a_y}\,{b_y} + {a_z}{b_z}\]
\[\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{a_x}}&{{a_y}}&{{a_z}} \\
  {{b_x}}&{{b_y}}&{{b_z}}
\end{array}} \right| = \hat i({b_z}{a_y} - {a_z}{b_y}) - \hat j({a_x}{b_z} - {b_x}{a_z}) + \hat k({a_x}{b_y} - {b_x}{a_y})\]

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.