Parallel and Perpendicular Vectors in Two Dimensions - Pre-Calculus

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Question

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

Answer

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 $\theta=136.2^\circ$

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Question

Find the dot product of the two vectors

and

.

Answer

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.

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

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Question

Find the dot product of the two vectors

and

.

Answer

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.

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Question

Find the dot product of the two vectors

and

.

Answer

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.

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

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Question

Let

Find the dot product of the two vectors

$u_1\bullet u_2$ .

Answer

Let

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

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

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Question

Let

Find the dot product of the two vectors

$u_1\bullet u_2$ .

Answer

Let

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

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

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Question

Evaluate the dot product of the following two vectors:

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

Answer

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:

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

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

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Question

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

Answer

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.

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

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Question

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.

,

Answer

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:

$\cos^{-1}\frac{a \cdot b}{\left | a\right |\left | b\right |}=\theta$

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

$\theta=6.667^o$

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Question

Find the angle between the two vectors:

Answer

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.

$\theta=147.41^o$

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Question

Find the measure of the angle between the following vectors:

Answer

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:

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

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

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

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

$\theta=cos^{-1}\left(\frac{-9}{\sqrt{1845}}\right)=102.09^{\circ}$

So the angle between these two vectors is 102.09 degrees

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Question

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

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

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

Answer

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.

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Question

Which pair of vectors represents two parallel vectors?

Answer

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.

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

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.

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Question

Find the measure of the angle between the two vectors.

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

Answer

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:

$\theta=\cos^{-1}{\frac{u_1v_1+u_2v_2+...+u_nv_n}{\left | u\right |\left | v\right |}}$

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.

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Question

Which of the following pairs of vectors are perpendicular?

Answer

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

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

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

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

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Question

Which of the following pairs of vectors are perpendicular?

Answer

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,

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

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

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Question

Which of the following pairs of vectors are perpendicular?

Answer

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

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

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Question

Which of the following pairs of vectors are perpendicular?

Answer

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,

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

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

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

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Question

Which of the following vectors are perpendicular?

Answer

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,

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

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

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Question

Which of the following pairs of vectors are perpendicular?

Answer

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,

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

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

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

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