91Ó°ÊÓ

The two vectors $\stackrel{â‡\P Ä}{a}\mathrm{and}\stackrel{â‡\P Ä}{b}$are given in such a way,

• role="math" $\stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c}\mathrm{and}a+b=c$
  
• $\stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{a}-\stackrel{â‡\P Ä}{b}$
  
• $\stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c}\mathrm{and}{a}^2+b^2=c^2$
  

The problem involvesthe law of vector algebra which includes arithmetic operations such as addition, subtraction, and multiplication on vectors. Here, the laws of vector algebra can be used to find out above mentioned vectors.

Formula:

$$\begin{gathered} \stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c} \\ \stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{b}+\stackrel{â‡\P Ä}{a} \end{gathered}$$

When the two vectors $\stackrel{â‡\P Ä}{a}\mathrm{and}\stackrel{â‡\P Ä}{b}$are parallel, then the sum of the magnitudes of two vectors is equal to the magnitude of the sum of the two vectors.

Consider $\stackrel{â‡\P Ä}{a}=5\hat{\mathrm{i}}\mathrm{and}\stackrel{â‡\P Ä}{b}=4\hat{\mathrm{i}}$are acting along the same direction as the x-axis. The magnitudes are$a=5\mathrm{and}b=4$

The sum of the magnitude of two vectors:

role="math" localid="1660890923071" $$\begin{gathered} a+b=c \\ 5+4=c \\ c=9 \end{gathered}$$

The magnitude of the sum of two vectors, according to the vector addition law,

$$\begin{gathered} \stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c} \\ 5\hat{i}+4\hat{i}=\stackrel{â‡\P Ä}{c} \\ 9\hat{i}=\stackrel{â‡\P Ä}{c} \\ c=9 \end{gathered}$$

Hence, when two vectors are acting in the same direction, then $\stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c}\mathrm{and}a+b=c$.

Vector$-\stackrel{â‡\P Ä}{b}$has the same magnitude as$\stackrel{â‡\P Ä}{b}$but the direction is opposite. So if we consider$\stackrel{â‡\P Ä}{b}$any vector other than zero vector (null vector), we cannot satisfy $\stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{a}-\stackrel{â‡\P Ä}{b}$.

When we put $\stackrel{â‡\P Ä}{b}=0$, then

$$\begin{gathered} \stackrel{â‡\P Ä}{a}+0=\stackrel{â‡\P Ä}{a}-0 \\ \stackrel{â‡\P Ä}{a}=\stackrel{â‡\P Ä}{a} \end{gathered}$$

Hence, we get the vector satisfying the given condition.

For this condition, the two vectors should be perpendicular to each other. Consider$\stackrel{â‡\P Ä}{a}$along the x-axis and$\stackrel{â‡\P Ä}{b}$along the y-axis, they$\stackrel{â‡\P Ä}{c}$should be in the x-y plane.

According to the triangle law of vector addition, when two vectors are perpendicular to each other, then the sum of the square of the magnitude of each vector is equal to the square of the magnitude of their resultant.

Hence, when the two vectors are acting perpendicular to each other, then

$\stackrel{â‡\P Ä}{a}+\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c}and{a}^2+b^2=c^2$