Matrices and Vectors - Pre-Calculus

Card 1 of 744

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Find the unit vector that is in the same direction as the vector $\vec{v}= [3,6,1]$

Tap to reveal answer

Answer

To find the unit vector in the same direction as a vector, we divide it by its magnitude.

The magnitude of $\vec{v}$ is $\left | \vec{v} \right |= \sqrt{3^2+6^2+1^2}=\sqrt{46}$ .

We divide vector $\vec{v}$ by its magnitude to get the unit vector $\vec{u}_v$ :

$\vec{u}_v= \frac{\vec{v}}{\left | \vec{v} \right |}=\frac{1}{\sqrt{46}}\cdot[3,6,1]$

or

$\vec{u}_v= \left [ {\frac{3}{\sqrt{46}}},\frac{6}{\sqrt{46}} , \frac{1}{\sqrt{46}} \right ]$

All unit vectors have a magnitude of , so to verify we are correct:

$\left | \vec{u}_v \right | = \sqrt{ (\frac{3}{\sqrt{46}})^2 + (\frac{6}{\sqrt{46}})^2 + (\frac{9}{\sqrt{46}})^2 } = \sqrt{\frac{9}{46} + \frac{36}{46} + \frac{1}{46} } = 1$

← Didn't Know|Knew It →

Question

A unit vector has length .

Given the vector

find the unit vector in the same direction.

Tap to reveal answer

Answer

First, you must find the length of the vector. This is given by the equation:

$\sqrt{3^2+6^2+44^2+(-5)^2+12^2}=\left | v\right |$

Then, dividing the vector by its length gives the unit vector in the same direction.

$\frac{v}{\left | v\right |}=\frac{<3,6,44,-5,12>}{\sqrt{2150}}\ \ \approx<0.065,0.129,0.949,-0.108,0.259>$

← Didn't Know|Knew It →

Question

Put the vector $\small (1,3)$ in unit vector form.

Tap to reveal answer

Answer

To get the unit vector that is in the same direction as the original vector $\small v$ , we divide the vector by the magnitude of the vector.

For $\small (1,3)$ , the magnitude is:

$\sqrt{x^2+y^2}$

$\small \sqrt{1^2+3^2}=\sqrt{10}$ .

This means the unit vector in the same direction of $\small (1,3)$ is,

$\small \frac{1}{\sqrt{10}}(1,3)=\left(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right)$ .

← Didn't Know|Knew It →

Question

Find the unit vector of

$\left<3,,4\right>$ .

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left | v\right |=\sqrt{a^{2}+b^{2}}$ .

For this vector in the problem

$\left | v\right |=\sqrt{3^{2}+4^{2}}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$\left | v\right |=5$ .

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 3,4\right>}{5}$ .

As such,

$u,=,\left<\frac{3}{5},, \frac{4}{5}\right>$ .

← Didn't Know|Knew It →

Question

Find the unit vector of

$\left<5,,12\right>$

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left | v\right |=\sqrt{a^{2}+b^{2}}$

For this vector in the problem

$\left | v\right |=\sqrt{5^{2}+12^{2}}$

$=\sqrt{25+144}$

$=\sqrt{169}$

$\left | v\right |=13$

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 5,12\right>}{13}$

As such,

$u,=,\left<\frac{5}{13},, \frac{12}{13}\right>$

← Didn't Know|Knew It →

Question

Write a vector equation describing the line passing through P1 (1, 4) and parallel to the vector $\vec{a}$ = (3, 4).

Tap to reveal answer

Answer

First, draw the vector $\vec{a}$ = (3, 4); this is represented in red below. Then, plot the point P1 (1, 4), and draw a line (represented in blue) through it that is parallel to the vector $\vec{a}$ .

We must find the equation of line . For any point P2 (x, y) on , $\vec{P1P2}=(x-1,y-4)$ . Since $\vec{P1P2}$ is on line and is parallel to $\vec{a}$ , $\vec{P1P2}=t\vec{a}$ for some value of t. By substitution, we have . Therefore, the equation is a vector equation describing all of the points (x, y) on line parallel to $\vec{a}$ through P1 (1, 4).

← Didn't Know|Knew It →

Question

True or false: A line through P1 (x1, y1) that is parallel to the vector $\vec{a}=(a_1,a_2)$ is defined by the set of points such that $\vec{P_1P_2}=t\vec{a}$ for some real number t. Therefore, .

Tap to reveal answer

Answer

This is true. The independent variable in this equation is called a parameter.

← Didn't Know|Knew It →

Question

Find the parametric equations for a line parallel to $\vec{a}=(3,2)$ and passing through the point (0, 5).

Tap to reveal answer

Answer

A line through a point (x1,y1) that is parallel to the vector $\vec a$ = (a1, a2) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 3t

y = 5 + 2t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 $\theta=136.2^\circ$

← Didn't Know|Knew It →

Question

Find the dot product of the two vectors

and

.

Tap to reveal answer

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.

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

← Didn't Know|Knew It →

Question

Find the dot product of the two vectors

and

.

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

Find the dot product of the two vectors

and

.

Tap to reveal answer

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.

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

← Didn't Know|Knew It →

Question

Let

Find the dot product of the two vectors

$u_1\bullet u_2$ .

Tap to reveal answer

Answer

Let

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

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

← Didn't Know|Knew It →

Question

Let

Find the dot product of the two vectors

$u_1\bullet u_2$ .

Tap to reveal answer

Answer

Let

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

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

← Didn't Know|Knew It →

Question

Evaluate the dot product of the following two vectors:

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

Tap to reveal answer

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:

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

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

← Didn't Know|Knew It →

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.

,

Tap to reveal answer

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.

$\theta=6.667^o$

← Didn't Know|Knew It →

Question

Find the angle between the two vectors:

Tap to reveal answer

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.

$\theta=147.41^o$

← Didn't Know|Knew It →

Question

Find the measure of the angle between the following vectors:

Tap to reveal answer

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

← Didn't Know|Knew It →

Question

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

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

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

0

Knew It