We have the following equation that shows the relation between the dot product of two vectors, \(\displaystyle \vec{A} \cdot \vec{B}\), to the relative angle between them \(\displaystyle \theta\),

\(\displaystyle \frac{\vec{A} \cdot \vec{B}}{\left| \vec{A}\right| \left| \vec{B} \right|} = \cos \theta\). 

From this, we can see that the numerator \(\displaystyle \vec{A} \cdot \vec{B}\) will be \(\displaystyle 0\) whenever \(\displaystyle \cos \theta=0\).  

\(\displaystyle \cos \theta = 0\) for all odd-multiples of \(\displaystyle \frac{\pi}{2}\), which in one rotation, includes \(\displaystyle \frac{\pi}{2}=90 ^{\circ},3\frac{\pi}{2}=270^{\circ}\).

There is no correct way to compute the above. In order to take the dot product, the two vectors must have the same number of components. These vectors have \(\displaystyle 4\) and \(\displaystyle 2\) components respectively.

To computer the dot product, we multiply the values of common components together and sum their totals. The outcome is a scalar value, not a vector.

So we have

\(\displaystyle < t_1,0,5> \cdot < t_8 ,8, 4> = (t_1)(t_8)+(0)(8)+(5)(4) = t_1t_8+20\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=3\widehat{i}-1\widehat{j}+4\widehat{k}\) and \(\displaystyle \overrightarrow{b}=2\widehat{i}+6\widehat{j}-2\widehat{k}\)

The dot product can be found following the example above:

\(\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(3*2)+(-1*6)+(4*-2)=-8\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=4\widehat{i}+5\widehat{j}+2\widehat{k}\) and \(\displaystyle \overrightarrow{b}=3\widehat{i}-3\widehat{j}+3\widehat{k}\)

The dot product can be found following the example above:

\(\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(4*3)+(5*-3)+(2*3)=3\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=2\widehat{i}-2\widehat{j}+8\widehat{k}\) and \(\displaystyle \overrightarrow{b}=3\widehat{i}-1\widehat{j}+1\widehat{k}\)

The dot product can be found following the example above:

\(\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(2*3)+(-2*-1)+(8*1)=16\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=cos(x)\widehat{i}+xy\widehat{j}+sin(y)\widehat{k}\) and \(\displaystyle \overrightarrow{b}=2xy\widehat{i}+3x\widehat{j}+2y\widehat{k}\)

The dot product can be found following the example above:

\(\displaystyle {\overrightarrow{a}\cdot\overrightarrow{b}=(cos(x)*2xy)+(xy*3x)+(sin(y)*2y)=2xycos(x)+3x^2y+2ysin(y)}\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=7\widehat{i}-1\widehat{j}-2\widehat{k}\) and \(\displaystyle \overrightarrow{b}=2\widehat{i}-14\widehat{j}+3\widehat{k}\)

The dot product can be found following the example above:

\(\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(7*2)+(-1*-14)+(-2*3)=22\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=13\widehat{i}+5\widehat{j}+12\widehat{k}\) and \(\displaystyle \overrightarrow{b}=3\widehat{i}+4\widehat{j}+5\widehat{k}\)

The dot product can be found following the example above:

\(\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(13*3)+(5*4)+(12*5)=119\)

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.

\(\displaystyle \overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}\)

\(\displaystyle \overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}\)

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

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 \(\displaystyle \overrightarrow{a}=4\widehat{i}-2\widehat{j}\) and \(\displaystyle \overrightarrow{b}=2\widehat{i}-4\widehat{j}\)

The dot product can be found following the example above:

\(\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(4*2)+(-2*-4)=16\)