The dot product of a paired set of vectors can be found by summing up the individual products of the multiplications between matched directional vectors.

$\displaystyle \overrightarrow{a}=\begin{bmatrix} a_1\\a_2 \\ ...\\ a_n \end{bmatrix}$

$\displaystyle \overrightarrow{b}=\begin{bmatrix}b_1 \\b_2 \\...\\b_n \end{bmatrix}$

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3+...+a_nb_n$

Note that the dot product is a scalar value rather than a vector; there's no directional term.

Now considering our problem, we're given the vectors

$\displaystyle \overrightarrow{a}=\begin{bmatrix} -6\\8 \\3 \end{bmatrix}$ and $\displaystyle \overrightarrow{b}=\begin{bmatrix}-6 \\-2 \\7 \end{bmatrix}$

The dot product can be found following the example above:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=36-16+21=41$

The dot product of a paired set of vectors can be found by summing up the individual products of the multiplications between matched directional vectors.

$\displaystyle \overrightarrow{a}=\begin{bmatrix} a_1\\a_2 \\ ...\\ a_n \end{bmatrix}$

$\displaystyle \overrightarrow{b}=\begin{bmatrix}b_1 \\b_2 \\...\\b_n \end{bmatrix}$

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3+...+a_nb_n$

Note that the dot product is a scalar value rather than a vector; there's no directional term.

Now considering our problem, we're given the vectors

$\displaystyle \overrightarrow{a}=\begin{bmatrix} 4\\ -9\\-9 \end{bmatrix}$ and $\displaystyle \overrightarrow{b}=\begin{bmatrix} -7\\ 7\\9 \end{bmatrix}$

The dot product can be found following the example above:

$\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=-28-63-81=-172$

The correct answer is $\displaystyle t\sin(t)+2t\cos(t)+t^3\ln(t)$.

To compute the dot product, we take the corresponding components of each vector, and multiply them together. In this case, we have $\displaystyle (t)(\sin(t))+(2t)(\cos(t))+(t^3)(\ln(t))=t\sin(t)+2t\cos(t)+t^3\ln(t)$.

Note that taking the dot product of any two vectors will always return a scalar-valued expression (or just a simple scalar). There should be no vector brackets in your answer.

The dot product of two vectors is defined as:

$\displaystyle \mathbf{u} \cdot\mathbf{v}=u_xv_x+u_yv_y+u_zv_z$

For the given vectors, this is:

$\displaystyle \mathbf{u} \cdot\mathbf{v}=(\frac{2\sqrt{2}}{ {3}})(6\sqrt{2})+(7)(0)+(4\sqrt{3})(\frac{5}{2})$

$\displaystyle \mathbf{u} \cdot\mathbf{v}=8+0+10\sqrt{3}=25.32$

The magnitude of a vector is given by:

$\displaystyle |\mathbf{u}|=\sqrt{u_x^2+u_y^2+u_z^2}=\sqrt{6^2+3^2+(-4)^2}=\sqrt{36+9+16}=\sqrt{61}$

Write the dot product formula.

$\displaystyle a\cdot b = a_1b_1+a_2b_2$

Substitute the values of the vectors and solve.  The dot product will result in a number, not a vector.

$\displaystyle (-2)(-3) + (9)(7) = 6 +63 =69$

The dot product is:  $\displaystyle 69$

To find the length $\displaystyle \small ||v||$ of the vector $\displaystyle \small v=(1,\sqrt{3})$, we take the square root of the dot product $\displaystyle \small v\cdot v$:

$\displaystyle \small \small ||v||=\sqrt{v\cdot v}=\sqrt{1^2+\sqrt{3}^2}=\sqrt{4}=2$

To find the length $\displaystyle \small ||v||$ of the vector $\displaystyle \small \small v=(15,12)$, we take the square root of the dot product $\displaystyle \small v\cdot v$:

$\displaystyle \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{15^2+12^2}=\sqrt{225+144}=\sqrt{369}$

To find the length $\displaystyle \small ||v||$ of the vector $\displaystyle \small \small \small v=(5,5)$, we take the square root of the dot product $\displaystyle \small v\cdot v$:

$\displaystyle \small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{5^2+5^2}=\sqrt{25+25}=\sqrt{25\cdot 2}=5\sqrt{2}$

To find the length $\displaystyle \small ||v||$ of the vector $\displaystyle \small \small \small \small v=(3,-1)$, we take the square root of the dot product $\displaystyle \small v\cdot v$:

$\displaystyle \small \small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$