Question

For any non-zero, non collinear vectors \[\overrightarrow p \] and \[\overrightarrow q \] the value of \[[\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k\] is
A.\[0\]
B.\[2(\overrightarrow p \times \overrightarrow q )\]
C.\[\overrightarrow q \times \overrightarrow p \]
D.\[\overrightarrow p \times \overrightarrow q \]

Answer 1

Hint:Here we use the concept that both the given vectors are non collinear and non-zero so the cross product of the two vectors will not be zero. Assume the values of magnitude in the vectors as different variables and perform cross product of the two vectors. Then multiply each term with the given directions to find the sum.

Formula used:Cross product of two vectors \[\overrightarrow a = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k,\overrightarrow b = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k\]is given by solving the determinant \[\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{x_1}}&{{y_1}}&{{z_1}} \\
  {{x_2}}&{{y_2}}&{{z_2}}
\end{array}} \right| = ({y_1}{z_2} - {y_2}{z_1})\hat i - ({x_1}{z_2} - {x_2}{z_1})\hat j + ({x_1}{y_2} - {x_2}{y_1})\hat k\]
* Dot product of two vectors \[\overrightarrow a = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k,\overrightarrow b = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k\]is given as \[\overrightarrow a \bullet \overrightarrow b = ({x_1}\hat i + {y_1}\hat j + {z_1}\hat k)({x_2}\hat i + {y_2}\hat j + {z_2}\hat k)\]
\[\overrightarrow a \bullet \overrightarrow b = {x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2}\]
* The product\[\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1\]
* Two vectors are said to be collinear if their cross product is zero vector and one vector can be written as multiple of another vector.

Complete step-by-step answer:
We have to find the value of \[[\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k ………...… (1)\]
Let us assume two vectors \[\overrightarrow p = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k\] and \[\overrightarrow q = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k\]
Then using the method of cross product of two vectors we can write
\[ \Rightarrow \overrightarrow p \times \overrightarrow q = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{x_1}}&{{y_1}}&{{z_1}} \\
  {{x_2}}&{{y_2}}&{{z_2}}
\end{array}} \right|\]
Solve the determinant to obtain the vector
\[ \Rightarrow \overrightarrow p \times \overrightarrow q = ({y_1}{z_2} - {y_2}{z_1})\hat i - ({x_1}{z_2} - {x_2}{z_1})\hat j + ({x_1}{y_2} - {x_2}{y_1})\hat k\]
We know that the \[\hat i.\hat i = 1,\hat j.\hat j = 1,\hat k.\hat k = 1\]and \[\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0\]
Take the dot product of the vector \[\overrightarrow p \times \overrightarrow q \] with each vector \[\hat i,\hat j,\hat k\] separately.
\[ \Rightarrow \hat i \bullet (\overrightarrow p \times \overrightarrow q ) = ({y_1}{z_2} - {y_2}{z_1})\hat i.\hat i - ({x_1}{z_2} - {x_2}{z_1})\hat j.\hat i + ({x_1}{y_2} - {x_2}{y_1})\hat k.\hat i\]
\[ \Rightarrow \hat i \bullet (\overrightarrow p \times \overrightarrow q ) = ({y_1}{z_2} - {y_2}{z_1}) ……………...… (2)\]
\[ \Rightarrow \hat j \bullet (\overrightarrow p \times \overrightarrow q ) = ({y_1}{z_2} - {y_2}{z_1})\hat i.\hat j - ({x_1}{z_2} - {x_2}{z_1})\hat j.\hat j + ({x_1}{y_2} - {x_2}{y_1})\hat k.\hat j\]
\[ \Rightarrow \hat j \bullet (\overrightarrow p \times \overrightarrow q ) = - ({x_1}{z_2} - {x_2}{z_1}) ………..… (3)\]
\[ \Rightarrow \hat k \bullet (\overrightarrow p \times \overrightarrow q ) = ({y_1}{z_2} - {y_2}{z_1})\hat i.\hat k - ({x_1}{z_2} - {x_2}{z_1})\hat j.\hat k + ({x_1}{y_2} - {x_2}{y_1})\hat k.\hat k\]
\[ \Rightarrow \hat k \bullet (\overrightarrow p \times \overrightarrow q ) = ({x_1}{y_2} - {x_2}{y_1}) …………..… (4)\]
Substitute the values from equation (2) in equation (1)
\[ \Rightarrow [\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k = [{y_1}{z_2} - {y_2}{z_1}]\hat i + [ - ({x_1}{z_2} - {x_2}{z_1})]\hat j + [{x_1}{y_2} - {x_2}{y_1}]\hat k\]
Since, \[\overrightarrow p \times \overrightarrow q = ({y_1}{z_2} - {y_2}{z_1})\hat i - ({x_1}{z_2} - {x_2}{z_1})\hat j + ({x_1}{y_2} - {x_2}{y_1})\hat k\]
Substitute the value of RHS as \[\overrightarrow p \times \overrightarrow q \]
\[ \Rightarrow [\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k = \overrightarrow p \times \overrightarrow q \]

So, the correct answer is “Option D”.

Note:Students might make the mistake of multiplying the directions directly by bringing the directions out of the bracket.
\[ \Rightarrow [\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k = [\overrightarrow p \overrightarrow q ]\hat i.\hat i + [\overrightarrow p \overrightarrow q ]\hat j.\hat j + [\overrightarrow p \overrightarrow q ]\hat k.\hat k\]
Use \[\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1\]
\[ \Rightarrow [\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k = [\overrightarrow p \overrightarrow q ] + [\overrightarrow p \overrightarrow q ] + [\overrightarrow p \overrightarrow q ]\]
Add the same terms in RHS to get the value
\[ \Rightarrow [\hat i\overrightarrow p \overrightarrow q ]\hat i + [\hat j\overrightarrow p \overrightarrow q ]\hat j + [\hat k\overrightarrow p \overrightarrow q ]\hat k = 3[\overrightarrow p \overrightarrow q ]\]
This is a wrong procedure. Keep in mind that the direction that is given with the vectors \[\overrightarrow p \overrightarrow q \]is to be multiplied within the bracket first and after obtaining the final value of the bracket we can multiply it to the value outside the bracket.