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

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

{\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