Vector

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Given vector and , solve for .

Explanation

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:

Given vector and , solve for .

Explanation

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

Explanation

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 .

Explanation

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 unit vector of .

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

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

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

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

Explanation

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:

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

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

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

Find the unit vector of .

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

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

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

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

Explanation

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:

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

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

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

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

$(\vec{a}\cdot \vec{b})\times\vec{c}$

$(\vec{a}\times \vec{b})\times\vec{c}$

$(\vec{a}\times \vec{b})\cdot \vec{c}$

$(\vec{a}\cdot \vec{b})+(\vec{a}\cdot \vec{c})$

$(\vec{a}\times \vec{b})\cdot(\vec{a}\times \vec{c})$

Explanation

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

$(\vec{a}\cdot \vec{b})\times\vec{c}$

$(\vec{a}\times \vec{b})\times\vec{c}$

$(\vec{a}\times \vec{b})\cdot \vec{c}$

$(\vec{a}\cdot \vec{b})+(\vec{a}\cdot \vec{c})$

$(\vec{a}\times \vec{b})\cdot(\vec{a}\times \vec{c})$

Explanation

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.

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

Given vector and , solve for .