Question

suppose \\( \vec{u} \\) and \\( \vec{v} \\) are vectors in \\( \mathbb{r}^{n} \\) with \\( \\| \vec{u}\\|=2,\\| \vec{v}\\|=3 \\), and the angle between them is \\( \theta=\pi / 3 \\). what is the value of \\( (\vec{u}-\vec{v}) \cdot(2 \vec{u}+3 \vec{v}) \\)?
(a) 4.
(b) -4.
(c) 8.
(d) -8.
(e) -16.

Explanation:

Step1: Expand the dot - product

Use the distributive property of the dot - product \((\vec{a}+\vec{b})\cdot(\vec{c}+\vec{d})=\vec{a}\cdot\vec{c}+\vec{a}\cdot\vec{d}+\vec{b}\cdot\vec{c}+\vec{b}\cdot\vec{d}\).
\((\vec{u}-\vec{v})\cdot(2\vec{u}+3\vec{v}) = 2\vec{u}\cdot\vec{u}+3\vec{u}\cdot\vec{v}-2\vec{v}\cdot\vec{u}-3\vec{v}\cdot\vec{v}\)
Since \(\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}\), we have \((\vec{u}-\vec{v})\cdot(2\vec{u}+3\vec{v})=2\vec{u}\cdot\vec{u}+(3\vec{u}\cdot\vec{v}-2\vec{u}\cdot\vec{v})-3\vec{v}\cdot\vec{v}=2\|\vec{u}\|^{2}+\vec{u}\cdot\vec{v}-3\|\vec{v}\|^{2}\)

Step2: Calculate \(\vec{u}\cdot\vec{v}\)

The formula for the dot - product is \(\vec{a}\cdot\vec{b}=\|\vec{a}\|\|\vec{b}\|\cos\theta\).
Given \(\|\vec{u}\| = 2\), \(\|\vec{v}\| = 3\), and \(\theta=\frac{\pi}{3}\), then \(\vec{u}\cdot\vec{v}=\|\vec{u}\|\|\vec{v}\|\cos\theta=2\times3\times\cos\frac{\pi}{3}\)
Since \(\cos\frac{\pi}{3}=\frac{1}{2}\), we have \(\vec{u}\cdot\vec{v}=2\times3\times\frac{1}{2}=3\)

Step3: Substitute the values of \(\|\vec{u}\|^{2}\), \(\vec{u}\cdot\vec{v}\), and \(\|\vec{v}\|^{2}\)

We know that \(\|\vec{u}\|^{2}=2^{2} = 4\) and \(\|\vec{v}\|^{2}=3^{2}=9\)
Substitute into \(2\|\vec{u}\|^{2}+\vec{u}\cdot\vec{v}-3\|\vec{v}\|^{2}\):
\(2\times4 + 3-3\times9=8 + 3-27=-16\)

Answer:

E. \(-16\)