Question

Match vector operations between two vectors A and B in column 1 with angles between the two vectors in column 2:

Column-1	column-2
a- $$|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$$	E-$${45}^0$$
b-$$|\overrightarrow{A}\times \overrightarrow{B}|=\overrightarrow{A}.\overrightarrow{B}$$	F-$${30}^0$$
c-$$\overrightarrow{A}.\overrightarrow{B}={\displaystyle \frac{AB}{2}}$$	G-$${90}^0$$
d-$$|\overrightarrow{A}\times \overrightarrow{B}|={\displaystyle \frac{AB}{2}}$$	H-$${60}^0$$

1) a-e, b-g, c-f, d-h
2) a-g, b-e, c-h, d-f
3) a-g, b-f, c-e, d-h
4) a-e, b-g, c-h, d-f

Answer 1

Hint: We are given a problem involving vectors, and we are given vector operations and there is some angle given in the second column. We need to solve the vector operations step by step to arrive at a meaningful conclusion.

Complete step by step answer:
We take first option: $$|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$$
$$R=\sqrt{A{}^{\text{2}}+B{}^{\text{2}}+2AB\cos \phi }$$ , where A and B are two forces, R is their resultant and $$\phi$$ is the angle between them.
If the value of angle is $${90}^0$$then cos $${90}^0$$=0, thus, $$|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$$
So, a-g.
We take second option: $$|\overrightarrow{A}\times \overrightarrow{B}|=\overrightarrow{A}.\overrightarrow{B}$$
The magnitude of cross product of two vectors is equal to the magnitude of dot product of two vectors.
Dot product or scalar product of two vectors is given by $$\overrightarrow{A}.\overrightarrow{B}=AB\cos \alpha$$
Where $$\alpha$$is the angle between the two vectors. Now for cross product,
$$|\overrightarrow{A}\times \overrightarrow{B}|=AB\mathrm{Sin}\alpha$$
Thus, $$\overrightarrow{A}.\overrightarrow{B}=AB\cos \alpha$$=$$|\overrightarrow{A}\times \overrightarrow{B}|=AB\mathrm{Sin}\alpha$$
This is possible when the sin and cosine components are the same and this happens only when the angle is $${45}^0$$. So, b-e
We take third option: $$\overrightarrow{A}.\overrightarrow{B}={\displaystyle \frac{AB}{2}}$$
Dot product or scalar product of two vectors is given by $$\overrightarrow{A}.\overrightarrow{B}=AB\cos \alpha$$
Where $$\alpha$$is the angle between the two vectors. Thus, the condition will be satisfied only when the value of the cosine component is $${\displaystyle \frac{1}{2}}$$and this happens only when $$\cos {60}^0$$. Thus, c-h.
$$\Rightarrow$$a-g, b-e, c-h and the only option left is d-f

So, the correct answer is “Option 2”.

Note:
While taking either dot product or cross product we have to keep in mind we have to take the angle between the two original vectors. the value of cos $$\alpha$$ varies over the given interval. We know the domain of cos is from -1 to 1. It attains its minimum value, -1 and maximum value, +1. Also, the value of cos $$\alpha$$is equal to zero when $$\alpha$$is equal to $$90{}^{\circ }$$.