Question

Let \( \vec{u}, \vec{v} \) and \( \vec{w} \) be vectors such that \( \vec{u} + \vec{v} + \vec{w} = \vec{0} \). If \( |\vec{u}| = 3, |\vec{v}| = 4 \) and \( |\vec{w}| = 5 \) then \( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} = \)

• A. \( 0 \)
• B. \( -25 \)
• C. \( 25 \)
• D. \( 50 \)
• E. \( 47 \)

Hint:

This problem is essentially finding the sum of the products of sides in a triangle. Since \( 3, 4, 5 \) is a Pythagorean triplet, these vectors actually form a right-angled triangle.

Answer 1

The Correct Option is B

Concept: The magnitude of the sum of three vectors squared is given by the identity: \[ |\vec{u} + \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \] This identity allows us to find the sum of dot products when the individual magnitudes and the total resultant vector are known.

Step 1: Evaluate the square of the resultant vector.
Given \( \vec{u} + \vec{v} + \vec{w} = \vec{0} \), we have: \[ |\vec{u} + \vec{v} + \vec{w}|^2 = |\vec{0}|^2 = 0 \]

Step 2: Substitute the given magnitudes into the identity.
We are given \( |\vec{u}| = 3, |\vec{v}| = 4, \) and \( |\vec{w}| = 5 \). \[ 0 = (3)^2 + (4)^2 + (5)^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \] \[ 0 = 9 + 16 + 25 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \] \[ 0 = 50 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]

Step 3: Solve for the sum of dot products.
\[ 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = -50 \] \[ \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} = \frac{-50}{2} = -25 \]