Question

The sum of two vectors \[\vec{A}\] and \[\vec{B}\] is at a right angle to their difference then
A. \[A=B\]
B. \[A=2B\]
C. \[B=2A\]
D. \[\vec{A}\] and \[\vec{B}\] have the same direction

Answer 1

Hint:A vector is a two-dimensional object with a magnitude and a direction.A vector can be visualized geometrically as a directed line segment with an arrow indicating the direction and a length equal to the magnitude of the vector.

Complete step by step answer:
Since vectors have both magnitude and direction, it is not possible to simply add two vectors to obtain their sum. The addition of vectors is more difficult than the addition of scalars.Consider the following scenario: a car traveling 10 miles north and 10 miles south.The total distance traveled, in this case, is 20 miles, but the displacement is zero. The North and South displacements are vector quantities, and their opposite directions cancel each other out.

Vector Subtraction: the addition of two vectors is analogous to the subtraction of two vectors. Suppose \[\bar{a}\] is to be subtracted from \[\bar{b}\]. \[\bar{a}-\bar{b}\] can be said as the addition of the vectors \[\bar{a}\] and \[(-\bar{b})\].

Thus, the addition formula can be applied as:
\[\bar{a}-\bar{b}=\sqrt{{{a}^{2}}+{{b}^{2}}-2ab.\cos \theta }\]
\[(-\bar{b})\] is nothing but reversed \[\bar{b}\] in direction.
Thus, The difference between two vectors \[\vec{A}\] and \[\vec{B}\] is,
\[(\bar{A}+\bar{B}).(\bar{A}-\bar{B})=0\]
\[\Rightarrow \bar{A}.\bar{A}+\bar{A}.\bar{B}-\bar{A}.\bar{B}-\bar{B}.\bar{B}=0~~\text{ }\]
\[\Rightarrow \bar{A}.\bar{A}=\bar{B}.\bar{B}\]
\[\therefore a=b\] if \[a\],\[b\] are magnitudes.

Thus, the answer is option A.

Note:A vector can be multiplied by another vector, but it cannot be divided. There are two types of vector products that are commonly used in physics and engineering. Scalar multiplication of two vectors is one type of multiplication. Taking the scalar product of two vectors yields a number (a scalar), as the name implies. Work and energy relations are defined using scalar products.