Question

Let $ \overrightarrow a = 2\widehat i + \widehat j - 2\widehat k $ and $ \overrightarrow b = \widehat i + \widehat {j.} $ Let $ \widehat c $ be a vector such that $ \left| {\overrightarrow c - \overrightarrow a } \right| = 3.\left| {(\overrightarrow a \times \overrightarrow {b)} \times \overrightarrow c } \right| = 3 $ and the angle between $ \overrightarrow c $ and $ \overrightarrow a \times \overrightarrow b $ be $ 30^\circ . $ Then $ \overrightarrow a \cdot \overrightarrow c $ is equal to.
A. $ \dfrac{{25}}{8} $
B. $ 2 $
C. $ 5 $
D. $ \dfrac{1}{8} $

Answer 1

Hint: Remember and apply perfectly the laws of dot product and cross product. Here we will use cross-product. It can be defined as the vector which is the binary operation on two vectors in three dimensions with the suspended sine angle between the two. First find the magnitude of the individual vectors and then the cross product and find the correlation from the given values and the required ones.

Complete step-by-step answer:
Given that - $ \overrightarrow a = 2\widehat i + \widehat j - 2\widehat k $
Take its magnitude.
 $ \Rightarrow \left| a \right| = \sqrt {{{(2)}^2} + {{(1)}^2} + {{( - 2)}^2}} $
Simplify,
 $
   \Rightarrow \left| a \right| = \sqrt {4 + 1 + 4} \\
   \Rightarrow \left| a \right| = \sqrt 9 \\
   \Rightarrow \left| a \right| = 3 \;
  $
Similarly, take magnitude for the vector, “b”
 $ \overrightarrow b = \widehat i + \widehat {j.} $
 $
   \Rightarrow \left| b \right| = \sqrt {{{(1)}^2} + {{(1)}^2}} \\
   \Rightarrow \left| b \right| = \sqrt 2 \;
  $
Now, given that –
 $ \left| {c - a} \right| = 3 $
Take square on both the sides of the equation –
 $ {\left| {c - a} \right|^2} = {3^2} $
Simplify the above equation. Open the brackets by using the square of the difference formula.
 $ \Rightarrow {\left| c \right|^2} + {\left| a \right|^2} - 2c.a = 9{\text{ }}.....{\text{(A)}} $
Also, given that –
 $ \left| {(\overrightarrow a \times \overrightarrow {b)} \times \overrightarrow c } \right| = 3 $
Apply the property of the cross-product –
 $ \left| {a \times b} \right|\left| c \right|\sin 30^\circ = 3 $
Place the value of $ \sin 30^\circ = \dfrac{1}{2} $ in the above equation –
 $ \left| {a \times b} \right|\left| c \right|\dfrac{1}{2} = 3 $
When any term in the division changes its side, it goes to the numerator of the opposite side.
 $
  \left| {a \times b} \right|\left| c \right| = 3 \times 2 \\
   \Rightarrow \left| {a \times b} \right|\left| c \right| = 6 \;
  $
Make mode of “c” as the subject –
 $ \Rightarrow \left| c \right| = \dfrac{6}{{\left| {a \times b} \right|}} $ ..... (B)
Now, find $ \left| {a \times b} \right| $
 $ a \times b = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  2&1&{ - 2} \\
  1&1&0
\end{array}} \right| $
Open the determinant-
 $ \Rightarrow a \times b = 2\widehat i - 2\widehat j + \widehat k $
Take mode in the above equation –
 $ \Rightarrow \left| {a \times b} \right| = \sqrt {{2^2} + {2^2} + {1^2}} $
Simplification –
 $
   \Rightarrow \left| {a \times b} \right| = \sqrt 9 \\
   \Rightarrow \left| {a \times b} \right| = 3{\text{ }}....{\text{ (C)}} \;
  $
Place value of equation (c) in the equation (B)
 $
   \Rightarrow \left| c \right| = \dfrac{6}{3} \\
   \Rightarrow \left| c \right| = 2\,{\text{ }}...{\text{ (D)}} \;
  $
Place values in the equation (A)
 $ \Rightarrow 4 + 9 - 2c.a = 9{\text{ }} $
Same terms with the same sign cancel each other from both sides of the equation. Therefore, remove
 $ \Rightarrow 4 - 2c.a = 0{\text{ }} $
Make unknown the subject –
 $
   \Rightarrow 2c.a = 4 \\
   \Rightarrow c.a = 2{\text{ }} \;
  $
So, the correct answer is “Option B”.

Note: Be careful while opening the brackets and simplification. Be good in multiples and concepts of square and square-root. Always remember the values of the angles for sine, cosine for direct substitution.