Question

A vector $\overline {{P_1}} $ is along the positive x-axis. If its cross product with another vector $\overline {{P_2}} $ is zero, then $\overline {{P_2}} $ could be:
(A) $4\widehat j$
(B) $ - 4\widehat i$
(C) $(\widehat i + \widehat j)$
(D) $ - (\widehat i + \widehat j)$

Answer 1

Hint: The vector product of two vectors $\overrightarrow A $and $\overrightarrow B $is defined by $\overrightarrow A \times \overrightarrow B = \left| A \right|\left| B \right|\sin \theta \widehat n.$ Where $\widehat n$is the unit vector perpendicular to both A & B vectors and $\theta $ is the angle between them.

Complete step by step answer:
It is given that $\overrightarrow {{P_1}} $is along the positive x-axis.
When two vectors are parallel, they are multiple of each other. i.e. if $\overrightarrow {{a_1}} ||\overrightarrow {{b_1}} $ then $\overrightarrow {{a_1}} = k\overrightarrow {{b_1}} $
And also, $\widehat i$is a position vector along the x-axis.
By using the explanation written above, we can write
$ \Rightarrow \overrightarrow {{P_1}} = k\widehat i,$ where $k$is some positive real number.
If $\overrightarrow {{P_1}} = 0$then none of the options given above will make any sense. Because, no matter what the value of $\overrightarrow {{P_2}} $is, $\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0$
So, let us assume, $\overrightarrow {{P_1}} \ne 0$
Since, $\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = \left| {\overrightarrow {{P_1}} } \right|\left| {\overrightarrow {{P_2}} } \right|\sin \theta \widehat n$
$\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0 \Rightarrow \left| {\overrightarrow {{P_1}} } \right|\left| {\overrightarrow {{P_2}} } \right|\sin \theta \widehat n = 0$
$\left| {\overrightarrow {{P_1}} } \right|\left| {\overrightarrow {{P_2}} } \right|\sin \theta \widehat n = 0 \Rightarrow \sin \theta = 0$
Therefore, $\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0$only if the sine of angle between them is zero. i.e. $\sin \theta = 0$
$\sin \theta $can be zero only if $\theta = 0$or $\theta = {180^0}$
$ \Rightarrow \overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0$ only if $\overrightarrow {{P_2}} $is along positive x-axis or negative x-axis.
From the given options, we can observe that, only option (B) satisfies the condition of $\overrightarrow {{P_2}} $
Therefore, the correct answer is option (B) $ - 4\widehat i.$

Note: The vector product is a vector quantity. $\widehat n$represents the direction of the cross product, $\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} .$Direction of $\widehat n$is perpendicular to the plane containing the vectors $\overrightarrow {{P_1}} $ and $\overrightarrow {{P_2}} .$ In other words we can say that $\widehat n$is perpendicular to both, $\overrightarrow {{P_1}} $ and $\overrightarrow {{P_2}} $ i.e. $\widehat n \bot \overrightarrow {{P_1}} $ and $\widehat n \bot \overrightarrow {{P_2}} .$
Since, $\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} $ is a vector quantity
$\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} \ne \overrightarrow {{P_2}} \times \overrightarrow {{P_1}} $
Because even though their magnitude will be equal but their direction will be opposite to each other.Examples of a vector quantity are, displacement, velocity, acceleration, Force etc.