Question

answer this question for 25 points.
if I like your answer I may make it the brainliest answer

Answer 1

Answer:

(b) A is perpendicular to B

Answer 2

Topic - Vector Analysis

We must know some formulae before we solve the problem:

1. If the two vectors $\vec{a}$ and $\vec{b}$ perpendicular to each other, then their dot product will be zero $(0)$, i.e., $\vec{a}.\vec{b}=0.$
2. If $\theta$ be the angle between the two vectors $\vec{a}$ and $\vec{b}$, then we write $\vec{a}.\vec{b}=|\vec{a}|\:|\vec{b}|\:cos\theta.$
3. If $|\vec{a}|$ be the magnitude of $\vec{a}$, then $(\vec{a})^{2}=\vec{a}.\vec{a}=|\vec{a}|^{2}.$
4. Vector scalar products satisfy commutative property, i.e., $\vec{a}.\vec{b}=\vec{b}.\vec{a}.$.
5. For any vector $\vec{a}$, we can write $|\vec{a}|=a.$

Let us solve the given problem now.

Given, $\vec{C}=\vec{A}+\vec{B}$

Squaring both sides, we get

$\quad (\vec{C})^{2}=(\vec{A}+\vec{B})^{2}$

$\Rightarrow \vec{C}.\vec{C}=(\vec{A}+\vec{B}).(\vec{A}+\vec{B})$

$\Rightarrow |\vec{C}|^{2}=\vec{A}.\vec{A}+2\:\vec{A}.\vec{B}+\vec{B}.\vec{B}$

$\Rightarrow |\vec{C}|^{2}=|\vec{A}|^{2}+2\:\vec{A}.\vec{B}+|\vec{B}|^{2}$

$\Rightarrow C^{2}=A^{2}+B^{2}+2\:\vec{A}.\vec{B}$

$\Rightarrow C^{2}=C^{2}+2\:\vec{A}.\vec{B}\quad[\because C^{2}=A^{2}+B^{2}]$

$\Rightarrow \vec{A}.\vec{B}=0$

Thus we conclude that the given two vectors $\vec{A}$ and $\vec{B}$ are perpendicular to each other.

Answer: Option (b) is correct.