In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| -3\:v\right \|=\left | -3\right |\left \| v\right \|$

$\displaystyle =3\cdot\sqrt{0^2+2^2}=3\cdot\sqrt{0+4}=3\sqrt{4}=3\cdot2=6$

As for the direction of the vector, since $\displaystyle c< 0$, the resulting vector will be against the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| 1\:v\right \|=\left | 1\right |\left \| v\right \|$

$\displaystyle =1\cdot\sqrt{(-2)^2+(-4)^2}=1\cdot\sqrt{4+16}=1\sqrt{20}=2\sqrt{5}$

As for the direction of the vector, since $\displaystyle c>0$, the resulting vector will be in the same direction as the original vector $\displaystyle v$

.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| 2\:v\right \|=\left | 2\right |\left \| v\right \|$

$\displaystyle =2\cdot\sqrt{1^2+2^2}=2\cdot\sqrt{1+4}=2\sqrt{5}$

As for the direction of the vector, since $\displaystyle c>0$, the resulting vector will be in the same direction as the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| -4\:v\right \|=\left | -4\right |\left \| v\right \|$

$\displaystyle =4\cdot\sqrt{(-3)^2+(-1)^2}=4\cdot\sqrt{9+1}=4\sqrt{10}$

As for the direction of the vector, since $\displaystyle c< 0$, the resulting vector will be against the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| -9\:v\right \|=\left | -9\right |\left \| v\right \|$

$\displaystyle =9\cdot\sqrt{1^2+4^2}=9\cdot\sqrt{1+16}=9\sqrt{17}$

As for the direction of the vector, since $\displaystyle c< 0$, the resulting vector will be against the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| -2\:v\right \|=\left | -2\right |\left \| v\right \|$

$\displaystyle =2\cdot\sqrt{4^2+10^2}=2\cdot\sqrt{16+100}=2\sqrt{116}=4\sqrt{29}$

As for the direction of the vector, since $\displaystyle c< 0$, the resulting vector will be against the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| 10\:v\right \|=\left | 10\right |\left \| v\right \|$

$\displaystyle =10\cdot\sqrt{2^2+(-1)^2}=10\cdot\sqrt{4+1}=10\sqrt{5}$

As for the direction of the vector, since $\displaystyle c>0$, the resulting vector in the same direction as the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| 13\:v\right \|=\left | 13\right |\left \| v\right \|$

$\displaystyle =13\cdot\sqrt{0^2+6^2}=13\cdot\sqrt{0+36}=13\sqrt{36}=13\cdot6=78$

As for the direction of the vector, since $\displaystyle c>0$, the resulting vector will be in the same direction as the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| 11\:v\right \|=\left | 11\right |\left \| v\right \|$

$\displaystyle =11\cdot\sqrt{5^2+(-1)^2}=11\cdot\sqrt{25+1}=11\sqrt{26}$

As for the direction of the vector, since $\displaystyle c>0$, the resulting vector will be in the same direction as the original vector $\displaystyle v$.

See below for a picture.

In order to solve the first part of the problem, we need to remember how to take the magnitude of a vector and scalar.

$\displaystyle \left \| c\:v\right \|=\left | c\right |\:\left \| v\right \|$, where $\displaystyle c$ is a scalar.

Now lets calculate this.

$\displaystyle \left \| -12\:v\right \|=\left | -12\right |\left \| v\right \|$

$\displaystyle =12\cdot\sqrt{4^2+4^2}=12\cdot\sqrt{16+16}=12\sqrt{32}=48\sqrt{2}$

As for the direction of the vector, since $\displaystyle c< 0$, the resulting vector will be against the original vector $\displaystyle v$.

See below for a picture.