Animated Solution for Physics - Kinematics: Let $|{\mathbf{A}}_1|=3,|{\mathbf{A}}_2|=5$ and $|{\mathbf{A}}_1+{\mathbf{A}}_2|=5$. The value of $(2{\mathbf{A}}_1+3{\mathbf{A}}_2)\cdot (3{\mathbf{A}}_1-2{\mathbf{A}}_2)$ is

Vector algebra often looks intimidating with its bold letters and dot products, but at its core, it behaves very much like the standard algebra we've known for years. Let's embark on a journey to decode this problem step-by-step.

Imagine you are standing at the origin of a coordinate system. You draw a vector ${\mathbf{A}}_1$ with a length of $3$ units. Then, from the same origin, you draw another vector ${\mathbf{A}}_2$ with a length of $5$ units. The problem tells us a fascinating fact: the resultant vector, ${\mathbf{A}}_1+{\mathbf{A}}_2$, also has a length of exactly $5$ units.

This geometric setup forms a triangle with sides $3$, $5$, and $5$. Because two sides are equal, it's an isosceles triangle! But before we get lost in geometry, let's use the raw power of vector algebra to extract the hidden relationship between ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$.

The key to unlocking any vector expression is knowing the dot product of the base vectors. We can find this using the fundamental law of vector addition:

Let's substitute the magnitudes we were given:

Notice how the $25$ beautifully cancels out on both sides. This leaves us with:

Insight: The negative sign of the dot product is a physical indicator that the angle between the two vectors is obtuse (greater than $90^{\circ }$). They are pointing somewhat away from each other!

Now, let's face the main challenge. We need to evaluate the expression:

Don't let the vectors scare you. The dot product is distributive, just like regular multiplication. Let's expand the brackets carefully:

Remember two golden rules of dot products: 1. The dot product of a vector with itself is the square of its magnitude: $\mathbf{A}\cdot \mathbf{A}=|\mathbf{A}{|}^2$. 2. The dot product is commutative: ${\mathbf{A}}_1\cdot {\mathbf{A}}_2={\mathbf{A}}_2\cdot {\mathbf{A}}_1$.

Applying these rules, our expression simplifies elegantly:

We have successfully broken down the complex expression into fundamental pieces that we already know! Let's plug in $|{\mathbf{A}}_1|=3$, $|{\mathbf{A}}_2|=5$, and our hard-earned dot product ${\mathbf{A}}_1\cdot {\mathbf{A}}_2=-4.5$:

And there we have it! By systematically applying vector properties, we transformed an abstract expression into a simple arithmetic calculation. The final answer is $-118.5$.