Vector Addition And Vector Subtraction

1) $120^{\circ}$
2) $60^{\circ}$
3) $180^{\circ}$
4) $90^{\circ}$

Solution:

From the triangle law of vector addition

So using the concept,

$
\begin{aligned}
& R^2=(3 P)^2+(2 P)^2+2 \cdot(3 P) \cdot(2 P) \cdot \cos \theta \\
& R^2=13 P^2+12 P^2 \cos \theta ....... (1)
\end{aligned}
$

When the first force is doubled, the resultant is doubled
$
\begin{aligned}
& \text { So, }(2 R)^2=(6 P)^2+(2 P)^2+2 \cdot(6 P)(2 P) \cdot \cos \theta \\
& \Rightarrow 4 R^2=36 P^2+4 P^2+24 P^2 \cos \theta \\
& \Rightarrow R^2=10 P^2+6 P^2 \cos \theta ............(2)
\end{aligned}
$

Equating (1) and (2)
$
\begin{aligned}
& \Rightarrow 13 P^2+12 P^2 \cos \theta=10 P^2+6 P^2 \cos \theta \\
& \Rightarrow 3 P^2=-6 P^2 \cos \theta \\
& \Rightarrow \cos \theta=-\frac{1}{2} \text { or } \theta=120^{\circ}
\end{aligned}
$

Equating (1) and (2)

$\begin{aligned}
& \Rightarrow 13 P^2+12 P^2 \cos \theta=10 P^2+6 P^2 \cos \theta \\
& \Rightarrow 3 P^2=-6 P^2 \cos \theta \\
& \Rightarrow \cos \theta=-\frac{1}{2} \text { or } \theta=120^{\circ}
\end{aligned}$

Hence, the answer is the option (1).

Example 2: The ratio of maxium and minimum magnitudes of the resultant of two vectors \text { }|\vec{a}| \text { and }|\vec{b}| is 3:1, Now |\vec{a}|=

1) $\mid \vec{b}$
2) ($2 \mid \vec{b}$
3) $3 \mid \vec{b}$
4) $4|\vec{b}|$

Solution:

$
\begin{aligned}
& |\vec{a}+\vec{b}|=\sqrt{a^2+b^2+2 a b \cos \theta} \\
& |\vec{a}+\vec{b}|_{\max }=a+b{ }_{\text {when } \theta=0^{\circ}} \\
& |\vec{a}+\vec{b}|_{\min }=a-b \quad \text { when } \theta=180^{\circ}
\end{aligned}
$
here
$
\frac{a+b}{a-b}=\frac{3}{1}
$
or
$
a+b=3 a-3 b
$
or $2 a=4 b \Rightarrow a=2 b$
$
|\vec{a}|=2|\vec{b}|
$

Hence, the answer is option (2).

Example 3:

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. The magnitudes of $(\vec{A}+\vec{B})$ is 'n' times the magnitudes of $(\vec{A}-\vec{B})$. The angle between $\vec{A}$ and $\vec{B}$ is :

1) $
\sin ^{-1}\left[\frac{n-1}{n+1}\right]
$
2) $ \cos ^{-1}\left[\frac{n^2-1}{n^2+1}\right]
$
3) $ \sin ^{-1}\left[\frac{n^2-1}{n^2+1}\right]
$
4) $\cos ^{-1}\left[\frac{n-1}{n+1}\right]
$

Solution:

$\begin{aligned}
& A^2+B^2+2 A B \cos \theta=n^2\left(A^2+B^2-2 A B \cos \theta\right) \\
& \because|\vec{A}|=|\vec{B}| \\
& \therefore A^2+A^2+2 A^2 \cos \theta=n^2\left(A^2+A^2-2 A^2 \cos \theta\right) \\
& 2 A^2(1+\cos \theta)=2 A^2 n^2(1-\cos \theta) \\
& \cos \theta\left(1+n^2\right)=n^2-1 \\
& \cos \theta=\frac{n^2-1}{n^2+1} \\
& \theta=\cos ^{-1}\left(\frac{n^2-1}{n^2+1}\right)
\end{aligned}$

Hence, the answer is the Option (2)

Example 4:

The magnitude and direction of the resultant of two vectors $\vec{A}$ and $\vec{B}$ in terms of their magnitudes and angle $\theta$ between them is :

1) $
R=\sqrt{A^2+B^2-2 A B \cos \theta}, \tan \alpha=\frac{B \sin \theta}{A+B \cos \theta}
$
2) $
R=\sqrt{A^2+B^2+2 A B \cos \theta}, \tan \alpha=\frac{B \cos \theta}{A+B \sin \theta}
$
3) $
R=\sqrt{A^2+B^2-2 A B \cos \theta}, \tan \alpha=\frac{A \sin \theta}{B+A \sin \theta}
$
4) $
R=\sqrt{A^2+B^2+2 A B \cos \theta}, \tan \alpha=\frac{B \sin \theta}{A+B \cos \theta}
$

Solution:

In the Parallelogram law of vector addition

If two vectors are represented by both magnitude and direction by two adjacent sides of a parallelogram taken from the same point then their resultant is also represented by both magnitude and direction taken from the same point but by the diagonal of the parallelogram.

The figure given below represents the law of parallelogram vector Addition.

$\vec{R}=\vec{A}+\vec{B}$

Using Pythagorean theorem in triangle MOP-

$\begin{aligned}
& R^2=(B \sin \Theta)^2+(A+B \cos \Theta)^2 \\
\Rightarrow & R=\sqrt{A^2+B^2+2 A B \cos \Theta} \\
\Rightarrow & \tan \alpha=\frac{B \sin \Theta}{A+B \cos \Theta}
\end{aligned}$
Hence, the answer is option (4).

Example 5: Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle with each other. If the force Q is doubled, then their resultant also gets doubled. Then the angle (in degrees) is :

1) 0

2) 120

3) 90

4) 110

Solution:

Parallelogram law of vector addition -

If two vectors are represented by both magnitude and direction by two adjacent sides of a parallelogram taken from the same point then their resultant is also represented by both magnitude and direction taken from the same point but by the diagonal of the parallelogram.

- wherein

$
R^2=P^2+Q^2+2 P Q \cos \theta
$

Represents the law of parallelogram Vector Addition
Case 1
$
R^2=4 F^2+9 F^2+12 F^2 \cos \theta
$

Case 2
$
\begin{aligned}
& 4 R^2=4 F^2+36 F^2+24 F^2 \cos \theta \\
& 16 F^2+36 F^2+48 F^2 \cos \theta=4 F^2+36 F^2+24 F^2 \cos \theta \\
& \therefore 12 F^2=-24 F^2 \cos \theta \\
& \theta=120^{\circ}
\end{aligned}
$
Hence, the answer is option (2).

Summary

Understanding vector addition and subtraction is crucial for accurately analyzing physical phenomena involving directional quantities. By mastering the methods for combining vectors, whether in the same or different directions, one can effectively apply these concepts to solve complex problems in physics and engineering, utilizing the triangle and parallelogram laws of vector addition