91Ó°ÊÓ

$$\begin{gathered} a⃗=(4.00m)\stackrel{Áåœ}{i}-(3.00m)\stackrel{Áåœ}{j} \\ \stackrel{→}{b}=(6.00\mathrm{m})\mathrm{i}+(8.00\mathrm{m})j \end{gathered}$$

Find the magnitude of any vector by using its values and perform different vector operations using vector algebra. Also,find the angle between vectors and axis usingtrigonometry.

Formulae are as follow:

$\mathrm{Magnitude}\mathrm{of}\stackrel{→}{\mathrm{a}}\mathrm{is}\mathrm{a}\sqrt{{\left(a_x\right)}^2+{\left(a_y\right)}^2}$

$$\begin{gathered} \mathrm{Angle}\mathrm{between}\stackrel{→}{a}\mathrm{and}\mathrm{X}\mathrm{axis}\mathrm{is}\mathrm{³Ù²¹²Ôθ}=\left(\frac{{\mathrm{a}}_{\mathrm{y}}}{{\mathrm{a}}_{\mathrm{x}}}\right) \\ \stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}=\left({\mathrm{a}}_{\mathrm{x}}+{\mathrm{b}}_{\mathrm{x}}\right)\stackrel{Áåœ}{\mathrm{i}}+\left({\mathrm{a}}_{\mathrm{y}}+{\mathrm{b}}_{\mathrm{y}}\right)\stackrel{Áåœ}{\mathrm{j}} \\ \stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{b}}=\left({\mathrm{a}}_{\mathrm{x}}-{\mathrm{b}}_{\mathrm{x}}\right)\stackrel{Áåœ}{\mathrm{i}}+\left({\mathrm{a}}_{\mathrm{y}}+{\mathrm{b}}_{\mathrm{y}}\right)\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$

where,a,b are vectors and $\mathrm{θ}$is the angle between$a_{\mathrm{x}}\mathrm{and}{a}_y$.

The magnitude of$\stackrel{→}{a}$ is,

$$\begin{gathered} a=\sqrt{{\left(4.0\right)}^2+{\left(-3.0\right)}^2} \\ =5.0\mathrm{m} \end{gathered}$$

Hence, magnitude of is 5.0 m .

The angle between$\stackrel{→}{a}$and X axis is,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{y}{x}\right) \\ ={\tan }^{-1}\left(\frac{-3.0}{4.0}\right) \\ =-37° \end{gathered}$$

Hence, the vector is 37⁰ clockwise from X axis.

The magnitude of $\stackrel{→}{b}$is,

$$\begin{gathered} b=\sqrt{{\left(6.0\right)}^2+{\left(8.0\right)}^2} \\ =10.0\mathrm{m} \end{gathered}$$

Hence,magnitude$\stackrel{→}{b}$ of is 10.0m .

The angle between $\stackrel{→}{b}$and X axis is,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{y}{x}\right) \\ ={\tan }^{-1}\left(\frac{8.0}{6.0}\right) \\ =53° \end{gathered}$$

Hence, angle of $\stackrel{→}{b}$is $53°$.

Use vector addition law to find the sum$\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$

$$\begin{gathered} \stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}=\left(6.0+4.0\right)\stackrel{Áåœ}{i}+\left(\left(8.0\right)+\left(-3.0\right)\right)\stackrel{Áåœ}{j} \\ =\left(10\right)\stackrel{Áåœ}{i}+\left(5\right)\stackrel{Áåœ}{j} \end{gathered}$$

Magnitude of$\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$ is calculated as,

$$\begin{gathered} \left|\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}\right|=\sqrt{{10}^2+5.0^2} \\ =11.0 \end{gathered}$$

Hence,magnitude ofrole="math" localid="1656946159137" $\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$ is 11m .

Angle of $\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$with X axis is,

role="math" localid="1656946653705" $$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{5.0}{10.0}\right) \\ =27° \end{gathered}$$

Hence, angle of $\stackrel{→}{\mathrm{a}}+\stackrel{→}{\mathrm{b}}$is$27°$.

Use vector subtraction law to find the $\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}$

$$\begin{gathered} \stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}=\left(6.0-4.0\right)\stackrel{Áåœ}{i}+\left(8.0+3.0\right)\stackrel{Áåœ}{j} \\ =\left(2.0\right)\stackrel{Áåœ}{i}+\left(11.0\right)\stackrel{Áåœ}{j} \end{gathered}$$

The magnitude can be found as,

role="math" localid="1656946628062" $$\begin{gathered} \left|\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}\right|=\sqrt{2^2+{11}^2} \\ =11 \end{gathered}$$

Hence, magnitude of $\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}$ is 11m .

The angle of$\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}$ with X axis is,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{11.0}{2.0}\right) \\ =80° \end{gathered}$$

Hence,angle of $\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}$is$80°$ .

$$\begin{gathered} \stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{b}}=\left(4.0-6.0\right)\stackrel{Áåœ}{i}+\left(\left(-3.0\right)-\left(8.0\right)\right)\stackrel{Áåœ}{j} \\ =\left(-2.0\right)\stackrel{Áåœ}{i}+\left(-11\right)\stackrel{Áåœ}{j} \\ \mathrm{Magnituted}\mathrm{of}\stackrel{\mathit{→}}{a}\mathit{}\mathit{-}\stackrel{\mathit{→}}{b}\mathrm{is}, \\ \left|\stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{b}}\right|=\sqrt{-2^2+\left(-11\right)\stackrel{Áåœ}{\mathrm{j}}} \\ =11.0 \\ \mathrm{Hence},\mathrm{magnitude}\mathrm{of}\stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{b}}\mathrm{is}11\mathrm{m}. \end{gathered}$$

Angle of $\stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{b}}$with X axis is,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{11.0}{2.0}\right) \\ =80° \end{gathered}$$

Hence, angle of is $80°$.

Since, this vector lies in Quadrant III, its angle with positive X axis is 80⁰+180⁰ = 260⁰.

Since,$\stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{b}}=-1\left(\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}\right)$, they are directed opposite to each other.

Hence, the angle between$\stackrel{→}{b}-\stackrel{→}{a}$and $\stackrel{→}{\mathrm{a}}-\stackrel{→}{b}$is $180°$

Therefore,by using laws of vector algebra, add and subtract the vectors. It is also possible to find the magnitude of vectors and their direction with particular axis.