Q.Find 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.

The key idea is the parallelogram law of vector addition, which gives both the magnitude and direction of the resultant.

Step 1 — Magnitude of the resultant

Place the vectors tail-to-tail with angle $\theta$ between them. Completing the parallelogram, the diagonal represents the resultant $\vec{R}$. Using the law of cosines on the triangle formed by $\vec{A}$, $\vec{B}$, and $\vec{R}$:

Step 2 — Direction of the resultant …

Figure 3.10 shows the construction this derivation is built on: $OP$ and $OQ$ represent $\vec{A}$ and $\vec{B}$ at angle $\theta$ to each other, and the parallelogram's diagonal $OS$ represents the resultant $\vec{R} = \vec{A} + \vec{B}$. Dropping the perpendicular $SN$ onto the extended line $OP$ (meeting it at $N$, with $PM$ perpendicular to $OS$) turns the geometry into the right triangles used below to derive $R$'s magnitude and its direction $\alpha$ from $\vec{A}$.

When you add two vectors, you're combining their effects. The key insight is that vectors don't add like plain numbers — direction matters. If you walk 5 km east and then 5 km north, you end up 7.07 km northeast, not 10 km. That's the whole story in a nutshell.

The most natural way to add vectors is the head-to-tail method: place the tail of $\vec{B}$ at the head of $\vec{A}$, then draw the resultant $\vec{R}$ from the tail of $\vec{A}$ to the head of $\vec{B}$. This creates a triangle, and the problem reduces to solving that triangle.

1. Set up the triangle

Let $\vec{A}$ and $\vec{B}$ have magnitudes $A$ and $B$, with an angle $\theta$ between them. When you place them head-to-tail, the angle inside the triangle at the vertex where $\vec{B}$ starts is not $\theta$ — it's $180^\circ - \theta$. Why? Because $\theta$ is the angle between the vectors when they share a tail. Once you shift $\vec{B}$ to the head of $\vec{A}$, the interior angle becomes supplementary to $\theta$.

2. Find the magnitude using the law of cosines

In any triangle with sides $A$, $B$, and $R$, where $R$ is opposite the angle $(180^\circ - \theta)$, the law of cosines gives:

Since $\cos(180^\circ - \theta) = -\cos\theta$, this becomes:

This is the magnitude of the resultant. Notice the plus sign — it comes from the fact that when $\theta$ is small (vectors nearly aligned), $\cos\theta$ is large and positive, so $R$ is close to $A+B$. When $\theta = 90^\circ$, $\cos\theta = 0$, and you get the Pythagorean theorem: $R = \sqrt{A^2 + B^2}$. When $\theta = 180^\circ$ (opposite directions), $\cos\theta = -1$, and $R = |A - B|$, the minimum possible.

3. Find the direction …

Concept: Resultant of Two Vectors via Component Resolution

Method: Analytical (Resolve-and-Add) Method, Not the Law of Cosines

Rather than treating the head-to-tail figure as a triangle and invoking the law of cosines/sines, this method resolves $\vec B$ into a component along $\vec A$ and a component perpendicular to $\vec A$, then adds the parallel pieces and the perpendicular pieces separately -- the standard "analytical method" of vector addition, distinct from the geometric triangle-law route.

Steps

1. Set up an axis along $\vec A$. Let $\vec A$ point along a reference direction; the angle between $\vec A$ and $\vec B$ is $\theta$.

2. Resolve $\vec B$ into components along and perpendicular to $\vec A$:

3. Add the components along $\vec A$. $\vec A$ itself contributes $A$ (it is the axis), and $\vec B$ contributes $B\cos\theta$ along the same axis:

4. Add the components perpendicular to $\vec A$. Only $\vec B$ has a perpendicular piece (by construction, $\vec A$ has none along its own perpendicular):

5. Combine the two perpendicular pieces by Pythagoras to get the magnitude -- $R_\parallel$ and $R_\perp$ are, by construction, at right angles to each other, so the resultant's magnitude is:

(using $\cos^2\theta+\sin^2\theta=1$). …