Question

If $|\vec{a}| = \sqrt{26}$, $|\vec{b}| = 7$, $|\vec{a} \times \vec{b}| = 35$, find $\vec{a} \cdot \vec{b}$

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

Lagrange's Identity ($|A \times B|^2 + (A \cdot B)^2 = |A|^2|B|^2$) is just a fancy vector version of the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. It completely bypasses the need to ever calculate the angle $\theta$!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$.

Step 2: Formula Application:
$(35)^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{26})^2 (7)^2$.

Step 3: Explanation:
$1225 + (\vec{a} \cdot \vec{b})^2 = 26 \times 49$. $26 \times 49 = 1274$. $(\vec{a} \cdot \vec{b})^2 = 1274 - 1225 = 49$. $\vec{a} \cdot \vec{b} = \sqrt{49} = 7$.

Step 4: Final Answer:
The dot product $\vec{a} \cdot \vec{b}$ is 7.