AP Calculus BC - Dot Product

Given the following two vectors, $\overrightarrow{a}=14\widehat{i}-6\widehat{j}$ and $\overrightarrow{b}=-5\widehat{i}-3\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Explanation

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.

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

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

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

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 $\overrightarrow{a}=14\widehat{i}-6\widehat{j}$ and $\overrightarrow{b}=-5\widehat{i}-3\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(-70)+(18)=-52$

Find the length $\small ||v||$ of the vector $\small v=(1,\sqrt{3})$ .

$\small \small ||v||=2$

$\small \small \small ||v||=4$

$\small \small \small ||v||=1+\sqrt{3}$

$\small \small \small ||v||=\sqrt{1+\sqrt{3}}$

Explanation

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

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

Given the following two vectors, $\overrightarrow{a}=3\widehat{i}-1\widehat{j}+4\widehat{k}$ and $\overrightarrow{b}=2\widehat{i}+6\widehat{j}-2\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Explanation

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.

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

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

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

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

The dot product can be found following the example above:

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

Given the following two vectors, $\overrightarrow{a}=1\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}-2\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Explanation

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.

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

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

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

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 $\overrightarrow{a}=1\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}-2\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(15)+(-22)=-7$

Given the following two vectors, $\overrightarrow{a}=7\widehat{i}-2\widehat{j}$ and $\overrightarrow{b}=1\widehat{i}+3\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Explanation

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.

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

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

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

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 $\overrightarrow{a}=7\widehat{i}-2\widehat{j}$ and $\overrightarrow{b}=1\widehat{i}+3\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(7)+(-6)=1$

Find the dot product between the vectors $\left \langle 2,y,7\right \rangle$ and $\left \langle 3,4,-1\right \rangle$

Explanation

To find the dot product between two vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ , we apply the following formula:

$a\cdot b=(x_1*x_2)+(y_1*y_2)+(z_1*z_2)$

Using the vectors from the problem statement, we get

Find the dot product of the vectors $\left \langle 3,-5,7\right \rangle$ and $\left \langle -6,-7,-4\right \rangle$

Explanation

To find the dot product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you use the following formula

$a\cdot b=a_1b_1+a_2b_2+a_3b_3$

Using the vectors from the problem statement, we get

Find the dot product of the vectors $\left \langle 8,9,5\right \rangle$ and $\left \langle 0,2,-7\right \rangle$

Explanation

To find the dot product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we use the formula $a\cdot b=a_1b_1+a_2b_2+a_3b_3$

Using the vectors from the problem statement, we get

Find the dot product of the vectors $\left \langle 2,-4,1\right \rangle$ and $\left \langle -1,3,6\right \rangle$

Explanation

To find the dot product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we apply the formula $a\cdot b=a_1b_1+a_2b_2+a_3b_3$

Using the vectors from the problem statement, we get

Find the dot product between the vectors $\left \langle -5,0,2\right \rangle$ and $\left \langle 10,2,7\right \rangle$

Explanation

To find the dot product between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you apply the formula: