Question

If \[\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k,\overrightarrow b = p\widehat i + \widehat j + q\widehat k\] and\[\overrightarrow b \times \overrightarrow a = 0\] then
1) \[(p,q) = (2,3)\]
2) \[(p,q) = ( - 2, - 3)\]
3) \[(p,q) = (1,2)\]
4) \[(p,q) = ( - 1, - 2)\]

Answer 1

Hint: Use the basic formula for the cross product of 2 vectors \[\overrightarrow x \times \overrightarrow y = \left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  {{x_1}}&{{x_2}}&{{x_3}} \\
  {{y_1}}&{{y_2}}&{{y_3}}
\end{array}} \right|\]
and equate it to 0 to find the value.

Complete step-by-step answer:
Given, \[\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k,\overrightarrow b = p\widehat i + \widehat j + q\widehat k\] and \[\overrightarrow b \times \overrightarrow a = 0\]
We know cross product of any 2 vectors is given by
\[\overrightarrow x \times \overrightarrow y = \left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  {{x_1}}&{{x_2}}&{{x_3}} \\
  {{y_1}}&{{y_2}}&{{y_3}}
\end{array}} \right|\]where \[\overrightarrow x = {x_1}\widehat i + {x_2}\widehat j + {x_3}\widehat k\]and \[\overrightarrow y = {y_1}\widehat i + {y_2}\widehat j + {y_3}\widehat k\]
Now, \[\overrightarrow b \times \overrightarrow a = \left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  p&1&q \\
  2&1&3
\end{array}} \right| = 0\]
\[ \Rightarrow (3 - q)\widehat i - (3p - 2q)\widehat j + (p - 2)\widehat k = 0\widehat i + 0\widehat j + 0\widehat k\]
\[
   \Rightarrow 3 - q = 0 \\
   \Rightarrow q = 3 \\
\]
Also,
\[
   \Rightarrow p - 2 = 0 \\
   \Rightarrow p = 2 \\
\]
Therefore, \[(p,q) = (2,3)\]
Hence, option 1) \[(p,q) = (2,3)\] is correct.

Note: Whenever the cross product of 2 vectors is equal to 0 then it implies the cross product is equal to \[0\widehat i + 0\widehat j + 0\widehat k\], the zero vector