Question

If $ \vec u = \hat j + 4\hat k $ , $ \vec v = \hat i - 3\hat k $ , and $ \vec w = \cos \theta \hat i + \sin \theta \hat j $ are vectors in 3 – dimensional space, then, the maximum possible value of \[\left| {\vec u \times \vec v.\vec w} \right|\] is?
A) $ \sqrt {14} $
B) 5
C) 7
D) $ \sqrt {13} $

Answer 1

Hint: We need to find the determinant of the given quantity, we can ignore the type of product signs amongst them, then an equation of function will be obtained by the derivative of it we can find its maximum value.
 $ \tan \theta = \dfrac{P}{B},\sin \theta = \dfrac{P}{H},\cos \theta = \dfrac{B}{H} $
Where P is the perpendicular, B is base and H is hypotenuse.

Complete step-by-step answer:
We have been given three vectors: $ \vec u = \hat j + 4\hat k $ , $ \vec v = \hat i - 3\hat k $ and $ \vec w = \cos \theta \hat i + \sin \theta \hat j $ and we are required to find their determinant. We do not need to consider their sign of product (dot or cross).
The given vectors with all the unit vectors can be written as:
 $ \vec u = 0\hat i + \hat j + 4\hat k $ , $ \vec v = \hat i + 0\hat j - 3\hat k $ , $ \vec w = \cos \theta \hat i + \sin \theta \hat j + 0\hat k $
We can find the required determinant using the coefficients of the unit vectors.
 \[\left| {\vec u \times \vec v.\vec w} \right| = \left| {\begin{array}{*{20}{c}}
  0&1&4 \\
  1&0&{ - 3} \\
  {\cos \theta }&{\sin \theta }&0
\end{array}} \right|\]
Calculating this determinant:
\[
\Rightarrow \left| {\vec u \times \vec v.\vec w} \right| = \left[ {0\left( {0 \times 0 - ( - 3)\sin \theta } \right)} \right] - 1\left[ {1 \times 0 - ( - 3)\cos \theta } \right] + 4\left[ {1 \times \sin \theta + 0 \times \cos \theta } \right] \\
  \left| {\vec u \times \vec v.\vec w} \right| = - 3\cos \theta + 4\sin \theta \\
  \left| {\vec u \times \vec v.\vec w} \right| = 4\sin \theta - 3\cos \theta \\
 \]
Let this value be the function of angle $ \theta $ , then:
 $ f\left( \theta \right) = 4\sin \theta - 3\cos \theta $
The maximum value of any function is given by the value of its variable when its derivative is equal to zero.
The differentiation of function is:
 $
\Rightarrow f'\left( \theta \right) = 4\cos \theta + 3\sin \theta \\
  \left(
  \because f'\left( {\cos \theta } \right) = - \sin \theta \ , f'\left( {\sin \theta } \right) = \cos \theta \
  \right) \
  $
The value obtained when we equate it to 0 is:
 $ 4\cos \theta + 3\sin \theta = 0 $
Dividing both sides by $ \cos \theta $ , we get:
 $
  4 + 3\tan \theta = 0 \\
  3\tan \theta = - 4 \\
\Rightarrow \tan \theta = \dfrac{{ - 4}}{3} \\
  $
Now, $ \tan \theta = \dfrac{P}{B} $ , so:
Perpendicular (P) = -4
Base (B) = 3
Then according to the Pythagoras theorem:
 $
  {H^2} = {P^2} + {B^2} \\
   \Rightarrow {H^2} = {( - 4)^2} + {(3)^2} \\
   \Rightarrow {H^2} = 16 + 9 \\
   \Rightarrow {H^2} = 25 \\
   \Rightarrow \sqrt {{H^2}} = \sqrt {25} \\
   \Rightarrow H = 5 \\
  $
The value of sine and cosine of angles can be calculated as:
 $
  \sin \theta = \dfrac{P}{H} \\
\Rightarrow \sin \theta = \dfrac{{ - 4}}{5} \\
  \cos \theta = \dfrac{B}{H} \\
\Rightarrow \cos \theta = \dfrac{3}{5} \\
  $
Substituting these values in the function and finding the magnitude:
 $
  \left| {f\left( \theta \right)} \right| = \left| {4 \times \left( { - \dfrac{4}{5}} \right) - 3 \times \dfrac{3}{5}} \right| \\
  \left| {f\left( \theta \right)} \right| = 5 \\
  $
Therefore, the maximum possible value of \[\left| {\vec u \times \vec v.\vec w} \right|\] is 5 and the correct option is B.
So, the correct answer is “Option B”.

Note: Here, $ \hat i,\hat j,\hat k $ used are called unit vectors and represent the vector along x, y and z axis respectively.
When we have to find the maximum possible value of any function in one variable, we differentiate it and equate the same to zero so as to obtain the value of the variable and then it is substituted in the function which then gives the maximum value.