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 Question:
We are given the magnitudes of two vectors and the magnitude of their cross product. We must calculate their scalar dot product.

Step 2: Detailed Explanation:
There is a highly powerful and direct mathematical identity linking the dot product and cross product of any two vectors, known as Lagrange's Identity:
$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$
Let's plug in the values provided by the question:
$|\vec{a}| = \sqrt{26} \implies |\vec{a}|^2 = 26$
$|\vec{b}| = 7 \implies |\vec{b}|^2 = 49$
$|\vec{a} \times \vec{b}| = 35 \implies |\vec{a} \times \vec{b}|^2 = 35^2 = 1225$
Substitute these squared values into Lagrange's Identity:
$1225 + (\vec{a} \cdot \vec{b})^2 = 26 \times 49$
Calculate the product on the right side:
$26 \times 49 = 1274$
Now, isolate the dot product term:
$1225 + (\vec{a} \cdot \vec{b})^2 = 1274$
$(\vec{a} \cdot \vec{b})^2 = 1274 - 1225$
$(\vec{a} \cdot \vec{b})^2 = 49$
Take the square root of both sides:
$\vec{a} \cdot \vec{b} = \pm 7$
Since all the provided multiple-choice options are strictly positive, we select the positive root.
$\vec{a} \cdot \vec{b} = 7$

Step 3: Final Answer:
The dot product is 7, matching option (d).