**Triangle law of vector addition** states that *if two vectors in magnitude and direction starting from a point represents two sides of a triangle in same order, then, the third side of the triangle taken in reverse order represents the resultant magnitude and direction of the two vectors*.

From the parallelogram law of vector addition, we know that the diagonal of the parallelogram represents the resultant of the two vectors represented by its adjacent sides.

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be any two vectors represented by $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ respectively as shown in the figure given below. Then, $\overrightarrow{OQ}$ is said to be the sum of the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ and is written as

\[\overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{PQ} =\overrightarrow{a}+\overrightarrow{b}\]

Thus, the sum of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is the vector $\overrightarrow{OQ}$ determined by the diagonal of the parallelogram whose adjacent sides are represented by $\overrightarrow{OP}$ and $\overrightarrow{PQ}$. This is also known as the **triangle law of vector addition**.

Triangle law of vector addition states that *if two vectors acting simultaneously at a point are represented in magnitude and direction by two sides of a triangle taken in same order, then the resultant vector both in magnitude and direction is represented by the closing (or third) side of the triangle taken in reverse order*.

Consider two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ are represented by two sides $\overrightarrow{AB}$ and $\overrightarrow{BC}$ of a triangle $ABC$. Then, according to the triangle law of vector addition, the third side $\overrightarrow{AC}$ represents the resultant $\overrightarrow{R}$ of those two vectors. Let $\theta$ be the angle between those two vectors.

\[\therefore\overrightarrow{AC} =\overrightarrow{AB}+\overrightarrow{BC}\] \[\overrightarrow{R}=\overrightarrow{P}+ \overrightarrow{Q}\]

### Magnitude of Resultant $\overrightarrow{R}$

From $C$, draw $CM$ perpendicular to $AB$, produced upto $M$.

By using Pythagoras theorem in right angled $\Delta AMC$, \[AC^2 = AM^2 + MC^2\] \[AC^2 = (AB + BM)^2+ MC^2 \text{ __(1)}\]

\[\text{and,}\;\sin\theta=\frac{MC}{BC}\] \[\therefore MC=Q\sin\theta \text{ __(2)}\]

\[\text{Similarly,}\;\cos\theta=\frac{BM}{BC}\] \[\therefore BM=Q\cos\theta \text{ __(3)}\]

\[\text{From (1),}\;R^2 = (P + Q\cos\theta)^2+ (Q\sin\theta)^2\] \[R^2 = P^2+ 2PQ\cos\theta + Q^2\cos^2\theta + Q^2\sin^2\theta\] \[R^2= P^2+ 2PQ\cos\theta + Q^2 (\cos^2\theta + \sin^2\theta)\] \[R^2= P^2+ 2PQ\cos\theta + Q^2\] \[\therefore R^2 = \sqrt{P^2 + 2PQ\cos\theta + Q^2}\]

Let the angle of the resultant with $\overrightarrow{P}$ be $\beta$. \[\text{Then,}\;\tan\beta = \frac{MC}{AM}\] \[\tan\beta = \frac{Q\sin\theta}{P+Q\cos\theta}\] \[\therefore\beta = \tan^{-1}\left(\frac{Q\sin\theta}{P+Q\cos\theta}\right)\]

#### Special Cases

**Case I:**

When the vectors are parallel to each other, i.e. $\theta=0°$, \[R^2 = \sqrt{P^2 + 2PQcos0 + Q^2}\] \[R^2 = \sqrt{P^2 + 2PQ + Q^2}\] \[R^2 = \sqrt{(P + Q)^2}\] \[∴R = P + Q\;\text{(Maximum value of R)}\]

**Case II:**

When the vectors are perpendicular to each other, i.e. $\theta=90°$, \[R^2 = \sqrt{P^2+ 2PQcos90 + Q^2}\] \[\therefore R^2 = \sqrt{P^2 + Q^2}\]

**Case III:**

When the angle between the vectors $(\theta)$ is $180°$. \[\text{Then,}\; R^2 = \sqrt{P² + 2PQcos180 + Q²}\] \[R^2 = \sqrt{P^2 – 2PQ + Q^2}\] \[R^2 = \sqrt{(P – Q)^2}\] \[\therefore R = P – Q\;\text{(Minimum Value of R)}\]

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