Question

if \\(vec{p} = \

$$\begin{pmatrix} 1 \\\\ -1 \\\\ 3 \\end{pmatrix}$$

\\), then \\(hat{j} \cdot \vec{p} =\\)

Explanation:

Step1: Recall the dot product formula for vectors

The unit vector \(\hat{j}\) in the standard basis is \(

$$\begin{pmatrix}0\\1\\0\end{pmatrix}$$

\), and for two vectors \(\vec{a}=

$$\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$$

\) and \(\vec{b}=

$$\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$$

\), the dot product \(\vec{a}\cdot\vec{b}=a_1b_1 + a_2b_2 + a_3b_3\).

Step2: Calculate the dot product \(\hat{j}\cdot\vec{p}\)

Given \(\vec{p}=

$$\begin{pmatrix}1\\-1\\3\end{pmatrix}$$

\) and \(\hat{j}=

$$\begin{pmatrix}0\\1\\0\end{pmatrix}$$

\), we use the dot product formula:
\(\hat{j}\cdot\vec{p}=(0\times1)+(1\times(-1))+(0\times3)\)

Step3: Simplify the expression

First, calculate each term: \(0\times1 = 0\), \(1\times(-1)=-1\), \(0\times3 = 0\). Then sum them up: \(0+(-1)+0=-1\).

Answer:

\(-1\)