Question

Let \(\vec{e_1}, \vec{e_2}\) be two non-collinear unit vectors such that \(|\vec{e_1}+\vec{e_2}| = \sqrt{3}\). Then evaluate \[ (2\vec{e_1}-5\vec{e_2}) \cdot (3\vec{e_1}+\vec{e_2}). \]

• A. \(\frac{11}{2}\)
• B. \(-\frac{11}{2}\)
• C. \(\frac{9}{2}\)
• D. \(-\frac{9}{2}\)

Hint:

Use the formula \(|\vec{u}+\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u}\cdot\vec{v}\) to find the dot product.

Answer 1

The Correct Option is B

Step 1: Compute \(\vec{e_1}\cdot\vec{e_2}\) from magnitude.
\[ |\vec{e_1}+\vec{e_2}|^2 = \vec{e_1}\cdot \vec{e_1} + 2 \vec{e_1}\cdot \vec{e_2} + \vec{e_2}\cdot \vec{e_2} = 1 + 1 + 2\vec{e_1}\cdot\vec{e_2} = 2 + 2\vec{e_1}\cdot\vec{e_2} \] \[ 2 + 2\vec{e_1}\cdot\vec{e_2} = (\sqrt{3})^2 = 3 \Rightarrow \vec{e_1}\cdot \vec{e_2} = \frac{1}{2} \]

Step 2: Expand the dot product.
\[ (2\vec{e_1}-5\vec{e_2}) \cdot (3\vec{e_1}+\vec{e_2}) = 2\vec{e_1}\cdot 3\vec{e_1} + 2\vec{e_1}\cdot \vec{e_2} -5\vec{e_2}\cdot 3\vec{e_1} -5\vec{e_2}\cdot\vec{e_2} \]

Step 3: Substitute values.
\[ = 6(\vec{e_1}\cdot \vec{e_1}) + 2(\vec{e_1}\cdot \vec{e_2}) -15(\vec{e_1}\cdot\vec{e_2}) -5(\vec{e_2}\cdot \vec{e_2}) = 6 + 1 - 7.5 -5 \]

Step 4: Simplify.
\[ 6 + 1 - 7.5 -5 = -\frac{11}{2} \]

Step 5: Final conclusion.
\[ \boxed{-\frac{11}{2}} \]