Question

If $\theta$ is an obtuse angle between vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}| = 5, |\vec{b}| = 3$ and $|\vec{a} \times \vec{b}| = 5\sqrt{5}$ then $\vec{a} \cdot \vec{b} = \dots$

• A. 10
• B. -10
• C. 5
• D. -5

Hint:

Alternatively, use Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.
$(5\sqrt{5})^2 + (\vec{a} \cdot \vec{b})^2 = (25)(9) \implies 125 + x^2 = 225 \implies x^2 = 100 \implies x = \pm 10$. Choose the negative root because the angle is obtuse!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the magnitudes of two vectors and the magnitude of their cross product. We must find the exact value of their dot product, factoring in the crucial piece of information that the angle between them is obtuse.

Step 2: Key Formula or Approach:
1. The magnitude of a cross product is $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.
2. The dot product is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.
3. Use the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$. An obtuse angle implies $\cos \theta$ is negative.

Step 3: Detailed Explanation:
Use the cross product formula to find $\sin \theta$:
$$5\sqrt{5} = (5)(3) \sin \theta$$
$$5\sqrt{5} = 15 \sin \theta \implies \sin \theta = \frac{5\sqrt{5}}{15} = \frac{\sqrt{5}}{3}$$
Now, use the Pythagorean identity to find $\cos \theta$:
$$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{\sqrt{5}}{3}\right)^2$$
$$\cos^2 \theta = 1 - \frac{5}{9} = \frac{4}{9}$$
Taking the square root gives two possibilities:
$$\cos \theta = \pm \frac{2}{3}$$
The problem specifically states that $\theta$ is an obtuse angle (between $90^\circ$ and $180^\circ$). In the second quadrant, cosine is strictly negative.
Therefore, $\cos \theta = -2/3$.
Finally, calculate the dot product:
$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$$
$$\vec{a} \cdot \vec{b} = (5)(3) \left(-\frac{2}{3}\right)$$
$$\vec{a} \cdot \vec{b} = 15 \left(-\frac{2}{3}\right) = -10$$

Step 4: Final Answer:
The dot product is -10, matching option (b).