Question

When two vectors are perpendicular their cross product is zero?

Answer 1

Hint: First, we define the terms vector, cross product, and perpendicular and solve the given problem further.
Vector has the magnitude and direction, where magnitude is also known as the size.
The length of the line or arrow shows the magnitude and arrowhead points ($\overrightarrow a $) is the direction.
Cross product vectors are the two vectors that can be defined as a binary operation on two vectors in three-dimensional spaces.
The cross-vector product of the vector always equals the vector.
Perpendicular is the line and that will make the angle of ${90^0}$with one another line.

Complete step by step answer:
First, given that the two vectors are perpendicular to each other, we can say if the two vectors are perpendicular to each other then the vectors angle between them will be equals to the ${90^0}$
The cross product of the given each vector equals the product of their magnitudes and sine of the angle between them
The magnitudes of the vectors are written as in the form of $A \times B = \left| A \right|\left| B \right|\sin \theta $
Given that two vectors are perpendicular then we get $\theta = {90^0}$
Substitute the value of sine in the magnitudes of the vector, we get $A \times B = \left| A \right|\left| B \right|\sin \theta \Rightarrow \left| A \right|\left| B \right|\sin {90^0}$
Since as from the trigonometric formula, $\sin {90^0} = 1$ (since the first quadrant touches the circumference of any circle with the y-axis and becomes a fraction of $\dfrac{1}{1}$)
Thus, we get, $A \times B = \left| A \right|\left| B \right|\sin {90^0} \Rightarrow \left| A \right|\left| B \right|$
Therefore, when two given vectors are perpendicular then their cross product is not zero but the dot product is zero.

Note: The commutative property holds in the scalar$a.b = b.a$, and not hold in the vector products as $a \times b \ne b \times a$
Parallel lines will not intersect with any of the other lines, unlike the perpendicular lines.
Parallel lines are more likely the straight lines in the given plane.
Since $\sin {90^0} = 1$but $\cos {90^0} = 0$and $\tan {90^0} = \dfrac{{\sin {{90}^0}}}{{\cos {{90}^0}}} \Rightarrow \dfrac{1}{0} = \infty $
Also $\cot {90^0} = \dfrac{1}{{\tan {{90}^0}}} = \dfrac{{\cos {{90}^0}}}{{\sin {{90}^0}}} \Rightarrow \dfrac{0}{1} = 0$