Vector - Math

Card 0 of 56

Find the unit vector of .

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To solve for the unit vector, the following formula must be used:

$\frac{v}{||v||}$

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

$\left | v \right |=$\sqrt{25+144}$$

$\left | v \right |=$\sqrt{169}$$

$\left | v \right |=13$

unit vector:

$\frac{<5,-12>}{13}$

$=\frac{1}{13}$*<5,-12>$

$=<$\frac{5}{13}$,$\frac{-12}{13}$>$

Given vector and , solve for .

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To solve for , We need to first multiply into vector to find and multiply into vector to find ; then we need to subtract the components in the vector and the components together:

Let $\vec{a}, \vec{b},\vec{c}$ be vectors. All of the following are defined EXCEPT:

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The cross product of two vectors (represented by "x") requires two vectors and results in another vector. By contrast, the dot product (represented by " $\cdot$ ") between two vectors requires two vectors and results in a scalar, not a vector.

If we were to evaluate $(\vec{a}\cdot \vec{b})\times\vec{c}$ , we would first have to evaluate $\vec{a}\cdot\vec{b}$ , which would result in a scalar, because it is a dot product.

However, once we have a scalar value, we cannot calculate a cross product with another vector, because a cross product requires two vectors. For example, we cannot find the cross product between 4 and the vector <1, 2, 3>; the cross product is only defined for two vectors, not scalars.

The answer is $(\vec{a}\cdot \vec{b})\times\vec{c}$ .

Let $\vec{}$ $\vec{a}$ and $\vec{b}$ be the following vectors: $\vec{a}=\left \langle 1,2,1\right \rangle$ and $\vec{b}=\left \langle 2,3,-1\right \rangle$ . If $\theta$ is the acute angle between the vectors, then which of the following is equal to $\theta$ ?

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The cosine of the acute angle between two vectors is given by the following formula:

$\cos\theta=\frac{\vec{a}$\cdot \vec{b}}{\left | \vec{a}\right | \left | \vec{b} \right |}$ , where $\vec{a} \cdot \vec{b}$ represents the dot product of the two vectors, $\left | \vec{a} \right |$ is the magnitude of vector a, and $\left | \vec{b} \right |$ is the magnitude of vector b.

First, we will need to compute the dot product of the two vectors. Let's say we have two general vectors in space (three dimensions), $\vec{a}$ and $\vec{b}$ . Let the components of $\vec{a}$ be and the components of $\vec{b}$ be . Then the dot product $\vec{a} \cdot \vec{b}$ is defined as follows:

$\vec{a} \cdot $\vec{b}=a_{1}$$(b_{1}$$)+a_{2}$$(b_{2}$$)+a_{3}$$(b_{3}$)$ .

Going back to the original problem, we can use this definition to find the dot product of $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ .

$\vec{a} \cdot $\vec{b}=a_{1}$$(b_{1}$$)+a_{2}$$(b_{2}$$)+a_{3}$$(b_{3}$)$

The next two things we will need to compute are $\left | \vec{a} \right |$ and $\left | \vec{b} \right |$ .

Let the components of a general vector $\vec{a}$ be . Then $\left | \vec{a} \right |$ is defined as $\left | \vec{a}\right |=$\sqrt{(a_${1}$)^2$+(a_${2})^2$+(a_$3)^2$}$ .

Thus, if $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ , then

$\left | \vec{a}\right |=$\sqrt{(a_${1}$)^2$+(a_${2})^2$+(a_$3)^2$}$

and $\left | \vec{b}\right |=$\sqrt{(b_${1}$)^2$+(b_${2})^2$+(b_$3)^2$$}=$\sqrt{(-2)^2$$+(3)^2$$+(-1)^2$}$$

$=$\sqrt{4+9+1}$=$\sqrt{14}$$ .

Now, we put all of this information together to find the cosine of the angle between the two vectors.

$\cos\theta=\frac{\vec{a}$\cdot \vec{b}}{\left | \vec{a}\right | \left | \vec{b} \right |}=\frac{3}{\sqrt{6}$\cdot$\sqrt{14}$}$

We just need to simplify this.

$$\frac{3}{\sqrt{6}$\cdot$\sqrt{14}$}=\frac{3}{\sqrt{6\cdot14}$}=\frac{3}{\sqrt{2\cdot3\cdot2\cdot7}$}$

$=\frac{3}{2\sqrt{21}$}$ .

In order to get it completely simplified, we have to rationalize the denominator by multiplying the numerator and denominator by the sqare root of 21.

$$\frac{3}{2\sqrt{21}$}=\frac{3\cdot\sqrt{21}$}{2$\sqrt{21}$\cdot$\sqrt{21}$}=\frac{3\sqrt{21}$}{2(21)}$

$=\frac{\sqrt{21}$}{2\cdot7}=\frac{\sqrt{21}$}{14}$ .

We just have one more step. We need to solve for the value of the angle. In order to do this, we can take the inverse cosine of both sides of the equation.

$\cos\theta=\frac{\sqrt{21}$}{14}$

.

The answer is .

Find the magnitude of vector :

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To solve for the magnitude of a vector, we use the following formula:

$\left | v \right $|=$\sqrt{x^2$$+y^2$}$$

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

$\left | v \right |=$\sqrt{9+}$4$

$\left | v \right |=$\sqrt{13}$$

Given vector and , solve for .

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To solve for , we need to add the components in the vector and the components together:

Given vector and , solve for .

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To solve for , we need to subtract the components in the vector and the components together:

Is $\langle $\frac{5}{13}$,$\frac{-12}{13}$\rangle$ a unit vector?

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To verify where a vector is a unit vector, we must solve for its magnitude. If the magnitude is equal to 1 then the vector is a unit vector:

$<$\frac{5}{13}$,$\frac{-12}{13}$>$

$$\sqrt{\left(\frac{25}{169}\right)+\left($\frac{144}{169}\right)$

$$\sqrt{\frac{169}{169}$}$

$\sqrt{1}$

$<$\frac{5}{13}$,$\frac{-12}{13}$>$ is a unit vector because magnitude is equal to .

Given vector . Solve for the direction (angle) of the vector:

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To solve for the direction of a vector, we use the following formula:

$tan\alpha$ = $\frac{b}{a}$

with the vector being $u=\left \langle a,b \right \rangle$

$u=\left \langle 3,3 \right \rangle$

$\tan\alpha=\frac{3}{3}$=1$

$\alpha = \arctan (1)$

$\alpha $=45^{\circ}$$

Solve for vector given direction of and magnitude of .

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To solve for a vector with the magnitude and direction given, we use the following formula:

$v=(8\cos $45^{\circ}$)i $+(8\sin45^{\circ}$)j$

$v=4$\sqrt{2}$i+4$\sqrt{2}$j$

$v=\left \langle 4$\sqrt{2}$,4$\sqrt{2}$} \right \rangle$

Given vector and , solve for .

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To solve for , We need to multiply into vector to find ; then we need to subtract the components in the vector and the components together:

Find the magnitude of .

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therefore the vector is

To solve for the magnitude:

$\sqrt{(36)+(36)}$

$\sqrt{72}$

$6$\sqrt{2}$$

Find the vector where its initial point is and its terminal point is .

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We need to subtract the -coordinate and the -coordinates to solve for a vector when given its initial and terminal coordinates:

Initial pt:

Terminal pt:

Vector:

Vector:

Find the vector where its initial point is and its terminal point is .

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We need to subtract the -coordinate and the -coordinate to solve for a vector when given its initial and terminal coordinates:

Initial pt:

Terminal pt:

Vector: $\left \langle 2-(-1),3-(-5)\right \rangle$

Vector: $\left \langle3,8\right \rangle$

Find the unit vector of .

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To solve for the unit vector, the following formula must be used:

$\frac{v}{||v||}$

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

$\left | v \right |=$\sqrt{25+144}$$

$\left | v \right |=$\sqrt{169}$$

$\left | v \right |=13$

unit vector:

$\frac{<5,-12>}{13}$

$=\frac{1}{13}$*<5,-12>$

$=<$\frac{5}{13}$,$\frac{-12}{13}$>$

Given vector and , solve for .

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To solve for , We need to first multiply into vector to find and multiply into vector to find ; then we need to subtract the components in the vector and the components together:

Let $\vec{a}, \vec{b},\vec{c}$ be vectors. All of the following are defined EXCEPT:

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The cross product of two vectors (represented by "x") requires two vectors and results in another vector. By contrast, the dot product (represented by " $\cdot$ ") between two vectors requires two vectors and results in a scalar, not a vector.

If we were to evaluate $(\vec{a}\cdot \vec{b})\times\vec{c}$ , we would first have to evaluate $\vec{a}\cdot\vec{b}$ , which would result in a scalar, because it is a dot product.

However, once we have a scalar value, we cannot calculate a cross product with another vector, because a cross product requires two vectors. For example, we cannot find the cross product between 4 and the vector <1, 2, 3>; the cross product is only defined for two vectors, not scalars.

The answer is $(\vec{a}\cdot \vec{b})\times\vec{c}$ .

Let $\vec{}$ $\vec{a}$ and $\vec{b}$ be the following vectors: $\vec{a}=\left \langle 1,2,1\right \rangle$ and $\vec{b}=\left \langle 2,3,-1\right \rangle$ . If $\theta$ is the acute angle between the vectors, then which of the following is equal to $\theta$ ?

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The cosine of the acute angle between two vectors is given by the following formula:

$\cos\theta=\frac{\vec{a}$\cdot \vec{b}}{\left | \vec{a}\right | \left | \vec{b} \right |}$ , where $\vec{a} \cdot \vec{b}$ represents the dot product of the two vectors, $\left | \vec{a} \right |$ is the magnitude of vector a, and $\left | \vec{b} \right |$ is the magnitude of vector b.

First, we will need to compute the dot product of the two vectors. Let's say we have two general vectors in space (three dimensions), $\vec{a}$ and $\vec{b}$ . Let the components of $\vec{a}$ be and the components of $\vec{b}$ be . Then the dot product $\vec{a} \cdot \vec{b}$ is defined as follows:

$\vec{a} \cdot $\vec{b}=a_{1}$$(b_{1}$$)+a_{2}$$(b_{2}$$)+a_{3}$$(b_{3}$)$ .

Going back to the original problem, we can use this definition to find the dot product of $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ .

$\vec{a} \cdot $\vec{b}=a_{1}$$(b_{1}$$)+a_{2}$$(b_{2}$$)+a_{3}$$(b_{3}$)$

The next two things we will need to compute are $\left | \vec{a} \right |$ and $\left | \vec{b} \right |$ .

Let the components of a general vector $\vec{a}$ be . Then $\left | \vec{a} \right |$ is defined as $\left | \vec{a}\right |=$\sqrt{(a_${1}$)^2$+(a_${2})^2$+(a_$3)^2$}$ .

Thus, if $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ , then

$\left | \vec{a}\right |=$\sqrt{(a_${1}$)^2$+(a_${2})^2$+(a_$3)^2$}$

and $\left | \vec{b}\right |=$\sqrt{(b_${1}$)^2$+(b_${2})^2$+(b_$3)^2$$}=$\sqrt{(-2)^2$$+(3)^2$$+(-1)^2$}$$

$=$\sqrt{4+9+1}$=$\sqrt{14}$$ .

Now, we put all of this information together to find the cosine of the angle between the two vectors.

$\cos\theta=\frac{\vec{a}$\cdot \vec{b}}{\left | \vec{a}\right | \left | \vec{b} \right |}=\frac{3}{\sqrt{6}$\cdot$\sqrt{14}$}$

We just need to simplify this.

$$\frac{3}{\sqrt{6}$\cdot$\sqrt{14}$}=\frac{3}{\sqrt{6\cdot14}$}=\frac{3}{\sqrt{2\cdot3\cdot2\cdot7}$}$

$=\frac{3}{2\sqrt{21}$}$ .

In order to get it completely simplified, we have to rationalize the denominator by multiplying the numerator and denominator by the sqare root of 21.

$$\frac{3}{2\sqrt{21}$}=\frac{3\cdot\sqrt{21}$}{2$\sqrt{21}$\cdot$\sqrt{21}$}=\frac{3\sqrt{21}$}{2(21)}$

$=\frac{\sqrt{21}$}{2\cdot7}=\frac{\sqrt{21}$}{14}$ .

We just have one more step. We need to solve for the value of the angle. In order to do this, we can take the inverse cosine of both sides of the equation.

$\cos\theta=\frac{\sqrt{21}$}{14}$

.

The answer is .

Find the magnitude of vector :

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To solve for the magnitude of a vector, we use the following formula:

$\left | v \right $|=$\sqrt{x^2$$+y^2$}$$

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

$\left | v \right |=$\sqrt{9+}$4$

$\left | v \right |=$\sqrt{13}$$

Given vector and , solve for .

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To solve for , we need to add the components in the vector and the components together: