Animated Solution for Mathematics - Vector Algebra: Let $\Delta PQR$ be a triangle. Let $\vec{a}=\overrightarrow{QR},\vec{b}=\overrightarrow{RP}$ and $\vec{c}=\overrightarrow{PQ}$. If $|\vec{a}|=12,|\vec{b}|=4\sqrt{3},\vec{b}\cdot \vec{c}=24$, then which of the following is (are) true?

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a triangle.

Imagine you are standing at vertex $P$ of a triangle $\Delta PQR$. You walk along the sides: first from $P$ to $Q$, then $Q$ to $R$, and finally $R$ to $P$. You have returned exactly to where you started.

In the language of vectors, this displacement is zero. The problem defines our vectors as $\vec{a}=\overrightarrow{QR}$, $\vec{b}=\overrightarrow{RP}$, and $\vec{c}=\overrightarrow{PQ}$. Because they form a closed loop, the triangle law of vector addition dictates:

This simple, elegant equation is the master key that will unlock every single part of this problem.

Now, let us look at Option D, which asks for the dot product $\vec{a}\cdot \vec{b}$. We are given the magnitude $|\vec{b}|=4\sqrt{3}$ and the dot product $\vec{b}\cdot \vec{c}=24$.

We use the dot product as a projection tool. By taking the dot product of our master equation $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ with $\vec{b}$, we get:

Expanding this, we obtain:

Substituting our known values, we have $\vec{a}\cdot \vec{b}+(4\sqrt{3}{)}^2+24=0$. Since $(4\sqrt{3}{)}^2=48$, the equation becomes $\vec{a}\cdot \vec{b}+48+24=0$, which simplifies to:

Just like that, Option D is confirmed as true.

Next, we turn our attention to the magnitude of $\vec{c}$. From our loop equation, we can isolate $\vec{c}$ as $\vec{c}=-(\vec{a}+\vec{b})$.

To find the magnitude squared, we square both sides:

We know $|\vec{a}|=12$, $|\vec{b}{|}^2=48$, and we just found $\vec{a}\cdot \vec{b}=-72$. Plugging these in:

With $|\vec{c}{|}^2=48$, we can easily test Options A and B. Option A asks for $\frac{|\vec{c}{|}^2}{2}-|\vec{a}|$, which is $\frac{48}{2}-12=24-12=12$.

This is true! Option B, however, results in $24+12=36$, which is not 30. Thus, Option A is true, and Option B is false.

Finally, we face the intimidating Option C: $|\vec{a}\times \vec{b}+\vec{c}\times \vec{a}|$. Do not let the notation frighten you.

In any triangle where $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, there is a beautiful symmetry:

This means our expression is simply $|\vec{a}\times \vec{b}+\vec{a}\times \vec{b}|=2|\vec{a}\times \vec{b}|$. To calculate this, we use Lagrange's Identity:

Substituting our values:

Taking the square root, $|\vec{a}\times \vec{b}|=\sqrt{1728}=24\sqrt{3}$. Multiplying by 2, we get $48\sqrt{3}$.

Option C is true! We have navigated the geometry, the algebra, and the identities to find that A, C, and D are the correct choices.