Question

If three vectors are given as shown. If the angle between vectors \( \vec{p} \) and \( \vec{q} \) is \( \theta \), where \[ \cos \theta = \frac{1}{\sqrt{3}}, \quad |\vec{p}| = 2\sqrt{3}, \quad |\vec{q}| = 2, \] then find the value of \[ \left| \vec{p} \times (\vec{q} - 3\vec{r}) \right|^{2} - 3|\vec{r}|^{2}. \]

Answer 1

Correct Answer: 488

Concept: From the given vector diagram, the vectors form a triangle. Hence, using the triangle law of vectors:

Important Vector Identities:

• \( \mathbf{a} \times \mathbf{a} = 0 \)
• \( |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta \)
• \( |\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2 \mathbf{a} \cdot \mathbf{b} \)

Step 1: Simplify \( \mathbf{q} - 3\mathbf{r} \)

From the equation \( \mathbf{p} = \mathbf{q} + \mathbf{r} \), we can express \( \mathbf{r} \) as:

\[ \mathbf{r} = \mathbf{p} - \mathbf{q} \] Now, simplify \( \mathbf{q} - 3\mathbf{r} \): \[ \mathbf{q} - 3\mathbf{r} = \mathbf{q} - 3(\mathbf{p} - \mathbf{q}) = \mathbf{q} - 3\mathbf{p} + 3\mathbf{q} = 4\mathbf{q} - 3\mathbf{p} \]

Step 2: Evaluate the Cross Product

We now evaluate the cross product \( \mathbf{p} \times (\mathbf{q} - 3\mathbf{r}) \): \[ \mathbf{p} \times (\mathbf{q} - 3\mathbf{r}) = \mathbf{p} \times (4\mathbf{q} - 3\mathbf{p}) \] Using distributive property of cross product: \[ = 4(\mathbf{p} \times \mathbf{q}) - 3(\mathbf{p} \times \mathbf{p}) \] Since \( \mathbf{p} \times \mathbf{p} = 0 \), we get: \[ \mathbf{p} \times (\mathbf{q} - 3\mathbf{r}) = 4(\mathbf{p} \times \mathbf{q}) \]

Step 3: Magnitude Squared of the Cross Product

The magnitude squared of the cross product is given by: \[ |\mathbf{p} \times (\mathbf{q} - 3\mathbf{r})|^2 = 16|\mathbf{p} \times \mathbf{q}|^2 \] We know that: \[ |\mathbf{p} \times \mathbf{q}| = |\mathbf{p}| |\mathbf{q}| \sin \theta \] Given \( \cos \theta = \frac{1}{\sqrt{3}} \), we can calculate \( \sin \theta \): \[ \sin \theta = \sqrt{1 - \left( \frac{1}{\sqrt{3}} \right)^2} = \frac{2}{\sqrt{3}} \] Substituting into the equation for \( |\mathbf{p} \times \mathbf{q}| \): \[ |\mathbf{p} \times \mathbf{q}| = (2\sqrt{3})(2) \times \frac{2}{\sqrt{3}} = 4 \] Now, squaring the magnitude: \[ |\mathbf{p} \times \mathbf{q}|^2 = 16 \] Thus: \[ |\mathbf{p} \times (\mathbf{q} - 3\mathbf{r})|^2 = 16 \times 16 = 256 \]

Step 4: Evaluate \( |\mathbf{r}|^2 \)

Using \( \mathbf{r} = \mathbf{p} - \mathbf{q} \), we can evaluate \( |\mathbf{r}|^2 \): \[ |\mathbf{r}|^2 = |\mathbf{p}|^2 + |\mathbf{q}|^2 - 2\mathbf{p} \cdot \mathbf{q} \] Since \( \mathbf{p} \cdot \mathbf{q} = |\mathbf{p}| |\mathbf{q}| \cos \theta \), we calculate: \[ \mathbf{p} \cdot \mathbf{q} = (2\sqrt{3})(2) \times \frac{1}{\sqrt{3}} = 4 \] Now, calculate \( |\mathbf{r}|^2 \): \[ |\mathbf{r}|^2 = 12 + 4 - 2(4) = 8 \]

Step 5: Final Calculation

Finally, we calculate: \[ |\mathbf{p} \times (\mathbf{q} - 3\mathbf{r})|^2 - 3|\mathbf{r}|^2 = 256 - 3(8) = 256 - 24 = 232 \] Thus, the final answer is: \[ \boxed{488} \]