Parallel and Perpendicular Vectors in Two Dimensions - Pre-Calculus

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Find the dot product of the two vectors

and

.

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To find the dot product of two vectors,

find the product of the x-components,

find the product of the y-components,

and then find the sum.

Find the dot product of the two vectors

and

.

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To find the dot product of two vectors,

find the product of the x-components,

find the product of the y-components,

and then find the sum.

$1\cdot 5+1\cdot5=5+5=10$

Find the angle between the following two vectors in 3D space.

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We can relate the dot product, length of two vectors, and angle between them by the following formula:

$a\cdot b=|a| \cdot |b| \cdot \cos \theta$

So the dot product of $a \cdot b$

and

is the addition of each product of components:

$2 \cdot -3 + 9 \cdot -4 + -3 \cdot 8 = -6 + (-36) + (-24) = -66$

now the length of the vectors of a and b can be found using the formula for vector magnitude:

$|a| = $$\sqrt{2^2$+ $9^2$ + $(-3)^2$}$ = $\sqrt{94}$$

$|b| = $$\sqrt{(-3)^2$+ $(-4)^2$ + $8^2$}$ = $\sqrt{89}$$

So:

$-66 = \sqrt94 \cdot \sqrt 89 \cdot \cos\theta$

$\cos \theta = -0.72158$ hence $\theta=136.2^\circ$

Find the dot product of the two vectors

and

.

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To find the dot product of two vectors,

find the product of the x-components,

find the product of the y-components,

and then find the sum.

$1\cdot 0+0\cdot1=0$

Let

Find the dot product of the two vectors

$u_1\bullet u_2$ .

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Let

The dot product $u_1\bullet u_2$ is equal to

$x_1x_2+y_1y_2 = 1\cdot 1+1\cdot0 =1+0=1$ .

Let

Find the dot product of the two vectors

$u_1\bullet u_2$ .

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Let

The dot product $u_1\bullet u_2$ is equal to

$x_1x_2+y_1y_2 = 1\cdot 0+0\cdot 1 =0+0=0$ .

Evaluate the dot product of the following two vectors:

$\left \langle -7, 4,-3\right \rangle\cdot \left \langle 8,11,-12\right \rangle=$

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To find the dot product of two vectors, we multiply the corresponding terms of each vector and then add the results together, as expressed by the following formula:

$\left \langle a_1,a_2,a_3\right \rangle\cdot \left \langle b_1,b_2,b_3\right \rangle=a_1b_1+a_2b_2+a_3b_3$

$\left \langle -7, 4,-3\right \rangle\cdot \left \langle 8,11,-12\right \rangle=(-7)(8)+(4)(11)+(-3)(-12)=24$

Determine the dot product of $\left \langle 2,3\right \rangle$ and $\left \langle -2,-3\right \rangle$ .

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The value of the dot product will return a number. The formula for a dot product is:

$\left \langle a,b\right \rangle \cdot\left \langle c,d\right \rangle = ac+bd$

Use the formula to find the dot product for the given vectors.

$\left \langle 2,3\right \rangle \cdot\left \langle -2,-3\right \rangle =2(-2)+3(-3) =-4-9=-13$

The dot product may be used to determine the angle between two vectors.

Use the dot product to determine if the angle between the two vectors.

,

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First, we note that the dot product of two vectors is defined to be;

$a\cdot b=\left | a\right |\left | b\right |\cos\theta$ .

First, we find the left side of the dot product:

$a\cdot b= (5)(1)+(24)(3)=77$ .

Then we compute the lengths of the vectors:

$\begin{align} \left | a\right |&= $$\sqrt{5^2$$+24^2$}$=$\sqrt{601}$\ \left | b\right |&= $$\sqrt{1^2$$+3^2$}$=$\sqrt{10}$ \end{align}$ .

We can then solve the dot product formula for theta to get:

Substituting the values for the dot product and the lengths will give the correct answer.

Find the angle between the two vectors:

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Solving the dot product formula for the angle between the two vectors results in the equation $\arccos ($\frac{a\cdot b}{\left | a\right | \left | b\right |}$)=\theta$ .

If we call the vectors a and b, finding the dot product and the lengths of the vectors, then substituting them into the formula will give the correct angle.

$a \cdot b =(2)(4)+(-5)(21)=-97$

$\begin{align} \left | a\right $|&=$\sqrt{2^2$$+(-5)^2$}$=$\sqrt{29}$ \ \left | b\right $|&=$\sqrt{4^2$$+21^2$}$=$\sqrt{457}$ \end{align}$

Substituting the values correctly will give the correct answer.

Find the measure of the angle between the following vectors:

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To find the angle between two vectors, use the following formula:

$cos(\theta)=\frac{v\cdot w}{\left | v\right |\left | w\right |}$$

$v \cdot w$ is known as the dot product of two vectors. It is found via the following formula:

$v_i\cdot w_i+v_j\cdot w_j=v \cdot w$

The denominator of the fraction involves multiplying the magnitude of each vector. To find the magnitude of a vector, use the following formula:

$\left | v\right $|=$\sqrt{i^2$$+j^2$}$$

Now we have everything we need to find our answer. Use our given vectors:

$v \cdot w=(5\cdot 3)+(-4\cdot 6)=15-24=-9$

$\left | v\right $|=$\sqrt{5^2$$+(-4)^2$}$=$\sqrt{41}$$

$\left | w\right $|=$\sqrt{3^2$$+6^2$}$=$\sqrt{45}$$

$cos(\theta)=\frac{-9}{\sqrt{41}$$\sqrt{45}$}$

So the angle between these two vectors is 102.09 degrees

$\vec{v}=-4i+5j$

$\vec{w}=5i+4j$

Which of the following best explains whether the two vectors above are perpendicular or parallel?

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Two vectors are perpendicular if their dot product is zero, and parallel if their dot product is 1.

Take the dot product of our two vectors to find the answer:

$v_i\cdot w_i+v_j\cdot w_j=v \cdot w$

Using our given vectors:

$\vec{v}=-4i+5j$

$\vec{w}=5i+4j$

$v\cdot w=(-4\cdot 5)+(5\cdot 4)=-20+20=0$

Thus our two vectors are perpendicular.

Which pair of vectors represents two parallel vectors?

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Two vectors are parallel if their cross product is . This is the same thing as saying that the matrix consisting of both vectors has determinant zero.

$\ \vector{v_$1}&=\left<x_{1}$$,y_{1}\right>\ \vector{v_$2}&=\left<x_{2}$$,y_{2}\right>\ M =\begin{pmatrix} $x_{1}$& $y_{1}$\ $x_{2}$__MATH_EXPR_2__y_{2}$ $\end{pmatrix}\det(M)=x_{1}$$y_{2}$$-x_{2}$$y_{1}$=0\Rightarrow\vector{v_1}\parallel\vector{v_2}$

This is only true for the correct answer.

$6\cdot 1-2\cdot3=0$

In essence each vector is a scalar multiple of the other.

Find the measure of the angle between the two vectors.

$v=<12,-4.5> $\text{ }$u=<-3,21>$

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We use the dot product to find the angle between two vectors. The dot product has two formulas:

$u\cdot v=u_1v_1+u_2v_2+...+u_nv_n=\left | u\right |\left | v\right |\cos\theta$

We solve for the angle measure to find the computational formula:

Our vectors give dot products and lengths:

$u\cdot v=-130.5\ \left | u\right |=12.816\ \left | v\right |=21.213$

Substituting these values into the formula above will give the correct answer.

Which of the following pairs of vectors are perpendicular?

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Two vectors are perpendicular when their dot product equals to .

Recall how to find the dot product of two vectors $\left<v_1, v_2\right>$ and $\left<w_1, w_2\right>$

$\left<v_1, v_2\right>\cdot \left<w_1, w_2\right>=v_1w_1+v_2w_2$

The correct choice is

$u=\left<3, 4\right>$

$v=\left<-8, 6\right>$

$\left<3, 4\right>\cdot \left<-8, 6\right>=(3\times(-8))+4 \times 6=-24+24=0$

Which of the following pairs of vectors are perpendicular?

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Two vectors are perpendicular when their dot product equals to .

Recall how to find the dot product of two vectors $\left<v_1, v_2\right>$ and

$\left<v_1, v_2\right>\cdot \left<w_1, w_2\right>=v_1w_1+v_2w_2$ .

The correct choice is,

$a=\left<5, 10\right>,b=\left<6, -3\right>$ .

$\left<5, 10\right>\cdot \left<6, -3\right>=(5)(6)+(10)(-3)=30-30=0$

Which of the following pairs of vectors are perpendicular?

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Two vectors are perpendicular when their dot product equals to .

Recall how to find the dot product of two vectors $\left<v_1, v_2\right>$ and $\left<w_1, w_2\right>$

$\left<v_1, v_2\right>\cdot \left<w_1, w_2\right>=v_1w_1+v_2w_2$ .

The correct choice is $c=\left<2, 1\right>, d=\left<3, -6\right>$ .

$\left<2, 1\right>\cdot \left<3, -6\right>=(2)(3)+(1)(-6)=6-6=0$

Which of the following pairs of vectors are perpendicular?

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Two vectors are perpendicular when their dot product equals to .

Recall how to find the dot product of two vectors $\left<v_1, v_2\right>$ and $\left<w_1, w_2\right>$

$\left<v_1, v_2\right>\cdot \left<w_1, w_2\right>=v_1w_1+v_2w_2$ .

Recall that for a vector, $v\left<a,b\right>=ai+bj$

The correct answer is then,

$e=3\sqrt3i-j=\left<3\sqrt3, -1\right>$

$f=\sqrt3i+9j=\left<\sqrt3, 9\right>$

$\left<3\sqrt3, -1\right>\cdot\left<\sqrt3, -9\right>=(3\sqrt3)(\sqrt3)+(-1)(9)=9-9=0$

Which of the following vectors are perpendicular?

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Two vectors are perpendicular when their dot product equals to .

Recall how to find the dot product of two vectors $\left<v_1, v_2\right>$ and $\left<w_1, w_2\right>$

$\left<v_1, v_2\right>\cdot \left<w_1, w_2\right>=v_1w_1+v_2w_2$ .

The correct answer is then,

$g=\left<$\frac{\sqrt2}{2}$, 6\right>, h=\left<-\sqrt2, $\frac{1}{6}\right>$

$\left<$\frac{\sqrt2}{2}$, 6\right>\cdot\left<-\sqrt2, $\frac{1}{6}\right>=\left($\frac{\sqrt2}{2}\right)(-\sqrt2)+(6)\left($\frac{1}{6}\right)=-1+1=0$

Which of the following pairs of vectors are perpendicular?

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Two vectors are perpendicular when their dot product equals to .

Recall how to find the dot product of two vectors $\left<v_1, v_2\right>$ and $\left<w_1, w_2\right>$

$\left<v_1, v_2\right>\cdot\left<w_1, w_2\right>=v_1w_1+v_2w_2$ .

Recall that for a vector, $v\left<a,b\right>=ai+bj$

The correct answer is then,

$k=\frac{1}{2}$i-5j=\left<$\frac{1}{2}$, -5\right>$

$l=10i+j=\left<10, 1\right>$

$\left<$\frac{1}{2}$, -5\right>\cdo\left<10, 1\right>=\left($\frac{1}{2}\right)(10)+(-5)(-1)=5-5=0$