Q.A girl walks 4 km towards west, then she walks 3 km in a direction $30^\circ$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Using the triangle law of vector addition, the three vectors form a closed triangle. Applying the cosine rule to the triangle formed by $\vec{b}$ and $\vec{c}$ (with $\vec{a}$ as their resultant) gives $\cos\theta = \frac{1}{2}$, so the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$.

Add the $\hat i,\hat j,\hat k$ components of the three vectors separately.

$\vec a+\vec b+\vec c$:

$\hat i$: $1-2+1=0$

$\hat j$: $-2+4-6=-4$

$\hat k$: $1+5-7=-1$

Sum $= 0\hat i - 4\hat j - \hat k = -4\hat j - \hat k$.

Walking along all three sides of a triangle in order brings you back to the starting point, so the net vector displacement is zero.

Let the triangle have vertices $A$, $B$, $C$. Taking the sides in order as vectors: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA}$.

By the triangle law of vector addition, $\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$. Adding $\overrightarrow{CA}$ (which is $-\overrightarrow{AC}$):

$\overrightarrow{AC} + \overrightarrow{CA} = \overrightarrow{AC} - \overrightarrow{AC} = \vec0$

Vector addition is associative — the way three vectors are grouped while adding does not affect the result.

For any three vectors $\vec{a},\vec{b},\vec{c}$, the associative property of vector addition states:

$(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$

This means whether we first add $\vec{a}$ and $\vec{b}$ and then $\vec{c}$, or first add $\vec{b}$ and $\vec{c}$ and then $\vec{a}$, the resultant vector is the same.

Traversing a triangle's three sides in order returns you to the starting point, so the vector sum is zero.

Let the triangle have vertices $A,B,C$. Taking the sides in order as $\vec{AB},\vec{BC},\vec{CA}$:

$\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AC}-\vec{AC}=\vec{0}.$

Geometrically, walking along all three sides in sequence brings you back to where you started — net displacement zero.

Add/subtract the vectors component-by-component ($\hat i,\hat j,\hat k$ separately).

Given $\vec a=2\hat i+0\hat j+3\hat k$, $\vec b=\hat i+2\hat j-3\hat k$, $\vec c=0\hat i+3\hat j-4\hat k$.

$\vec a-\vec b=(2-1)\hat i+(0-2)\hat j+(3-(-3))\hat k=\hat i-2\hat j+6\hat k$.

$2\vec c=6\hat j-8\hat k$.

$\vec a-\vec b+2\vec c=(1+0)\hat i+(-2+6)\hat j+(6-8)\hat k=\hat i+4\hat j-2\hat k$.

Checking against the given options — (a) $2\hat i-4\hat j+8\hat k$, (b) $\hat i-4\hat j-8\hat k$, (c) $\hat i+4\hat j-8\hat k$, (d) $\hat i+4\hat j+8\hat k$ — none reproduces the correctly-computed $\hat k$ coefficient of $-2$, though (c) and (d) match the $\hat i,\hat j$ parts. This is most likely an error introduced when the options were extracted/typeset, not a genuine ambiguity in the vector algebra. The mathematically correct result is $\hat i+4\hat j-2\hat k$.

The statement is True — this is a direct consequence of the triangle law of vector addition.

Let the triangle have vertices $A,B,C$. Taking the sides in order means $\overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CA}$ (each side's vector starts where the previous one ended).

By the triangle law, $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$.

Adding $\overrightarrow{CA}$ to both sides:

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{AC}+\overrightarrow{CA}=\vec0$

(since $\overrightarrow{CA}=-\overrightarrow{AC}$).

Two forces at right angles combine like the legs of a right triangle — the resultant is the hypotenuse, found with the Pythagorean form of the parallelogram law.

When two forces (or vectors) $P$ and $Q$ act at a point with an angle $\theta$ between them, the magnitude of their resultant is:

$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$

Here $P=3$, $Q=4$, $\theta = 90^\circ$, so $\cos\theta = 0$:

$R = \sqrt{3^2 + 4^2 + 0} = \sqrt{9+16} = \sqrt{25} = 5$

definition recall

Triangle law of vectors: If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant (sum) is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Symbolically, if $\overrightarrow{AB}=\vec a$ and $\overrightarrow{BC}=\vec b$, then $\overrightarrow{AC}=\vec a+\vec b$.