Dot product has the form of
$u_1v_1+u_2v_2+...u_nv_n$
In this case, you can write the standard form as $x_1y_1+x_2y_2+...x_ny_n$
Since,
x = [0.5 0.5]
y = [0.75, 1.25]
$x_1 = 0.5 \space y_1 = 0.75$
$x_2 = 0.5 \space y_2 = 1.25$
Therefore, $x \cdot y$ is
$(0.5)(0.75) + (0.5)(1.25)$
$0.375 + 0.625 =1$
When we have to Transpose a Matrix, we need to change the order. For example, x = [0.5 0.5] is a 1 x 2 matrix because it has one row and two columns.
Therefore, $x^T$ would be a 2 x 1 matrix and would be written as $\left(\begin{array}{c} .5 \\ .5 \end{array}\right)$. We see two rows and one column and that's a good sign that we did it correctly.
For this problem, a $x^T$ form of a matrix is written as
$
\begin{array}{c}
x_1 = 0.5\\
x_2 =0.5\\
\vdots\\
x_n
\end{array}$
I've added the values so it would be easier to see.
Now $x^T$ multiply by $y$
$\left(\begin{array}{c} .5 \\ .5 \end{array}\right)\left(\begin{array}{cc} .75 & 1.25 \end{array}\right)$
would be $(0.5)(0.75) + (0.5)(1.25)$ = $0.375 + 0.625 = 1$
In general, if $\vec{x} = a\vec{i}+b\vec{j}$, then the magnitude of $\vec{x}$, denoted $\lVert \vec{x}\rVert,$ is given by $\lVert \vec{x}\rVert = \sqrt{a^2+b^2}$. Does this help?
Best Answer
A quick way to get it:
You know the lengths of the two vectors and that $ \ \vec{A} \bullet \vec{B} \ $ is given by $ || \vec{A} || \cdot || \vec{B} || \ \cos \theta \ $ . The magnitude of $ \ \vec{A} \ \times \ \vec{B} \ $ is $ || \vec{A} || \cdot || \vec{B} || \ \sin \theta \ , $ so find $ \ \cos \theta \ $ and use $ \ \sin \theta \ = \ \sqrt{1 - \cos^2 \theta} \ $ .