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Question: A vector \(\overline {{P_1}} \) is along the positive x-axis. If its cross product with another vect...

A vector P1‾\overline {{P_1}} is along the positive x-axis. If its cross product with another vector P2‾\overline {{P_2}} is zero, then P2‾\overline {{P_2}} could be:
(A) 4j^4\widehat j
(B) −4i^ - 4\widehat i
(C) (i^+j^)(\widehat i + \widehat j)
(D) −(i^+j^) - (\widehat i + \widehat j)

Explanation

Solution

The vector product of two vectors A→\overrightarrow A and B→\overrightarrow B is defined by A→×B→=∣A∣∣B∣sin⁡θn^.\overrightarrow A \times \overrightarrow B = \left| A \right|\left| B \right|\sin \theta \widehat n. Where n^\widehat nis the unit vector perpendicular to both A & B vectors and θ\theta is the angle between them.

Complete step by step answer:
It is given that P1→\overrightarrow {{P_1}} is along the positive x-axis.
When two vectors are parallel, they are multiple of each other. i.e. if a1→∣∣b1→\overrightarrow {{a_1}} ||\overrightarrow {{b_1}} then a1→=kb1→\overrightarrow {{a_1}} = k\overrightarrow {{b_1}}
And also, i^\widehat iis a position vector along the x-axis.
By using the explanation written above, we can write
⇒P1→=ki^,\Rightarrow \overrightarrow {{P_1}} = k\widehat i, where kkis some positive real number.
If P1→=0\overrightarrow {{P_1}} = 0then none of the options given above will make any sense. Because, no matter what the value of P2→\overrightarrow {{P_2}} is, P1→×P2→=0\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0
So, let us assume, P1→≠0\overrightarrow {{P_1}} \ne 0
Since, P1→×P2→=∣P1→∣∣P2→∣sin⁡θn^\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = \left| {\overrightarrow {{P_1}} } \right|\left| {\overrightarrow {{P_2}} } \right|\sin \theta \widehat n
P1→×P2→=0⇒∣P1→∣∣P2→∣sin⁡θn^=0\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0 \Rightarrow \left| {\overrightarrow {{P_1}} } \right|\left| {\overrightarrow {{P_2}} } \right|\sin \theta \widehat n = 0
∣P1→∣∣P2→∣sin⁡θn^=0⇒sin⁡θ=0\left| {\overrightarrow {{P_1}} } \right|\left| {\overrightarrow {{P_2}} } \right|\sin \theta \widehat n = 0 \Rightarrow \sin \theta = 0
Therefore, P1→×P2→=0\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0only if the sine of angle between them is zero. i.e. sin⁡θ=0\sin \theta = 0
sin⁡θ\sin \theta can be zero only if θ=0\theta = 0or θ=1800\theta = {180^0}
⇒P1→×P2→=0\Rightarrow \overrightarrow {{P_1}} \times \overrightarrow {{P_2}} = 0 only if P2→\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 P2→\overrightarrow {{P_2}}
Therefore, the correct answer is option (B) −4i^. - 4\widehat i.

Note: The vector product is a vector quantity. n^\widehat nrepresents the direction of the cross product, P1→×P2→.\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} .Direction of n^\widehat nis perpendicular to the plane containing the vectors P1→\overrightarrow {{P_1}} and P2→.\overrightarrow {{P_2}} . In other words we can say that n^\widehat nis perpendicular to both, P1→\overrightarrow {{P_1}} and P2→\overrightarrow {{P_2}} i.e. n^⊥P1→\widehat n \bot \overrightarrow {{P_1}} and n^⊥P2→.\widehat n \bot \overrightarrow {{P_2}} .
Since, P1→×P2→\overrightarrow {{P_1}} \times \overrightarrow {{P_2}} is a vector quantity
P1→×P2→≠P2→×P1→\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.