Question

If \[\mathop A\limits^ \to \times \mathop B\limits^ \to = \mathop C\limits^ \to + \mathop D\limits^ \to \] , then select the correct alternative
 $ \left( a \right){\text{ }}\mathop A\limits^ \to {\text{ is parallel to }}\mathop C\limits^ \to + \mathop D\limits^ \to $
 $ \left( b \right){\text{ }}\mathop A\limits^ \to {\text{ is perpendicular to }}\mathop C\limits^ \to $
 $ \left( c \right){\text{ Component of }}\mathop C\limits^ \to {\text{ along }}\mathop A\limits^ \to {\text{ - component of }}\mathop A\limits^ \to {\text{ along }}\mathop D\limits^ \to $
 $ \left( d \right){\text{ Component of }}\mathop C\limits^ \to {\text{ along }}\mathop A\limits^ \to {\text{ = - component of }}\mathop D\limits^ \to {\text{ along }}\mathop A\limits^ \to $

Answer 1

Hint: As we know that the cross product of two vectors will always point in a direction that will be perpendicular to the plane. And from this, we can say that the vector will be at the right angle to both the vectors. By using this we can answer this question.

Complete step-by-step answer:
Here we have the vector equation given as \[\mathop A\limits^ \to \times \mathop B\limits^ \to = \mathop C\limits^ \to + \mathop D\limits^ \to \] and we have to choose the correct alternatives.
As we know that the according to the definition of cross product the vector \[\mathop C\limits^ \to + \mathop D\limits^ \to \] will be perpendicular to both the vector which is \[\mathop A\limits^ \to \] and \[\mathop B\limits^ \to \] .
So mathematically it can be written as
 $ \Rightarrow \mathop A\limits^ \to \cdot \left( {\mathop C\limits^ \to + \mathop D\limits^ \to } \right) = 0 $
Here, we can see that there is the dot product and on solving the dot product for the above equation, we will get the equation as
 $ \Rightarrow \mathop A\limits^ \to \cdot \mathop C\limits^ \to + \mathop A\limits^ \to \cdot \mathop D\limits^ \to = 0 $
Here, we can say that
 $ \Rightarrow {\text{A}}\left( {{\text{Component of }}\mathop C\limits^ \to {\text{ along }}\mathop A\limits^ \to } \right){\text{ + A}}\left( {{\text{component of }}\mathop D\limits^ \to {\text{ along }}\mathop A\limits^ \to } \right) = 0 $
Or on solving taking the one braces to the right side, we get the above line as
 $ \Rightarrow {\text{A}}\left( {{\text{Component of }}\mathop C\limits^ \to {\text{ along }}\mathop A\limits^ \to } \right){\text{ = - A}}\left( {{\text{component of }}\mathop D\limits^ \to {\text{ along }}\mathop A\limits^ \to } \right) = 0 $
Hence, the option $ \left( d \right) $ is correct.
So, the correct answer is “Option D”.

Note: In a cross product, the magnitude of the cross product will be equal to the area of a parallelogram and which would lie on the same plane as a vector. In the cross product, the direction can be obtained by using the right-hand thumb rule. So if our index finger is the first vector and the middle finger would be the second vector then the thumb finger will be the cross product of those vectors.
For solving this question we just need to know the basics of the cross and dot products. So let’s see the application of it. For the practical application of the cross product, it will be the torque. We can see in our daily life as all the screws and bottle caps are designed in a way that when they move upwards, at that time it gets rotated anticlockwise and vice-versa.