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: Find \(\vec e_1\cdot\vec e_2\) from the given magnitude.
\[ |\vec e_1+\vec e_2|^2 = |\vec e_1|^2+2\vec e_1\cdot\vec e_2+|\vec e_2|^2 = 1+2\vec e_1\cdot\vec e_2+1 = 3 \] Solving, \(\vec e_1\cdot\vec e_2 = \frac{1}{2}\).

Step 2: Expand the required dot product, collecting like terms first.
\[ (2\vec e_1-5\vec e_2)\cdot(3\vec e_1+\vec e_2) = 6|\vec e_1|^2 + (2-15)(\vec e_1\cdot\vec e_2) - 5|\vec e_2|^2 \]

Step 3: Substitute the known magnitudes and dot product only once, at the end.
\[ = 6(1) - 13\left(\frac{1}{2}\right) - 5(1) = 6 - \frac{13}{2} - 5 \]

Step 4: Combine into a single fraction.
\[ = 1 - \frac{13}{2} = \frac{2-13}{2} = -\frac{11}{2} \]
\[ \boxed{-\frac{11}{2}} \]