Let `PQR` be a triangle . Let `veca=overline(QR),vecb = overline(RP) and vecc= overline(PQ).if |veca|=12, |vecb|=4sqrt3and vecb.vecc= 24` then which of the following is (are) true ?

Options

A. (a) `|vecc|^(2)/2-|veca|=12`
B. (b) `|vecc|^(2)/2-|veca|=30`
C. (c) `|vecaxxvecb + veccxxveca|= 48sqrt3`
D. (d) `veca.vecb=-72`

Explanation

To solve the problem, we will analyze the triangle \( PQR \) and the vectors associated with its sides. We have:

- \( \vec{a} = \overline{QR} \) with magnitude \( |\vec{a}| = 12 \)
- \( \vec{b} = \overline{RP} \) with magnitude \( |\vec{b}| = 4\sqrt{3} \)
- \( \vec{c} = \overline{PQ} \) with the condition \( \vec{b} \cdot \vec{c} = 24 \)

We need to determine which of the given options are true based on these vectors.

### Step 1: Use the Law of Cosines
From the triangle, we can apply the law of cosines. We know that:

\[ |\vec{a}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2 |\vec{b}| |\vec{c}| \cos \theta \]

Where \( \theta \) is the angle opposite side \( \vec{a} \).

### Step 2: Substitute the Known Values
Substituting the known magnitudes into the equation:

\[ 12^2 = (4\sqrt{3})^2 + |\vec{c}|^2 - 2 (4\sqrt{3}) |\vec{c}| \cos \theta \]

Calculating \( 12^2 \) and \( (4\sqrt{3})^2 \):

\[ 144 = 48 + |\vec{c}|^2 - 8\sqrt{3} |\vec{c}| \cos \theta \]

### Step 3: Rearranging the Equation
Rearranging gives:

\[ |\vec{c}|^2 - 8\sqrt{3} |\vec{c}| \cos \theta + 48 - 144 = 0 \]

This simplifies to:

\[ |\vec{c}|^2 - 8\sqrt{3} |\vec{c}| \cos \theta - 96 = 0 \]

### Step 4: Use the Dot Product Condition
We also know from the problem that:

\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \phi = 24 \]

Where \( \phi \) is the angle between \( \vec{b} \) and \( \vec{c} \). Substituting \( |\vec{b}| \):

\[ (4\sqrt{3}) |\vec{c}| \cos \phi = 24 \]

This gives:

\[ |\vec{c}| \cos \phi = \frac{24}{4\sqrt{3}} = 2\sqrt{3} \]

### Step 5: Solve for \( |\vec{c}| \)
Now we can substitute \( |\vec{c}| \cos \phi \) back into our earlier equation. We can express \( \cos \theta \) in terms of \( \phi \) using the relationship between angles in a triangle:

\[ \cos \theta = -\cos \phi \]

Substituting \( \cos \phi \):

\[ |\vec{c}|^2 + 8\sqrt{3} |\vec{c}| \cdot 2\sqrt{3} - 96 = 0 \]

### Step 6: Solve the Quadratic Equation
This is a quadratic equation in terms of \( |\vec{c}| \):

\[ |\vec{c}|^2 + 48 |\vec{c}| - 96 = 0 \]

Using the quadratic formula:

\[ |\vec{c}| = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-48 \pm \sqrt{48^2 + 4 \cdot 96}}{2} \]

Calculating the discriminant:

\[ 48^2 + 384 = 2304 + 384 = 2688 \]

Thus:

\[ |\vec{c}| = \frac{-48 \pm \sqrt{2688}}{2} \]

Calculating \( \sqrt{2688} \):

\[ \sqrt{2688} = 52 \]

So:

\[ |\vec{c}| = \frac{-48 \pm 52}{2} \]

This gives two solutions:

1. \( |\vec{c}| = 2 \)
2. \( |\vec{c}| = -50 \) (not valid)

### Conclusion
Thus, we find \( |\vec{c}| = 2 \).