The dimensions of the vectors are not the same.  Placeholders cannot be added to a vector.  Therefore, the values of the vectors cannot be added.

The correct answer is:  $\displaystyle \textup{ Cannot be solved.}$

We find the norm of a vector by finding the sum of each element squared and then taking the square root.

$\displaystyle \left | \vec{v} \right |= \sqrt{2^2+4^2+3^2}=\sqrt{29}$.

We find the norm of a vector by finding the sum of each component squared and then taking the square root of that sum.

$\displaystyle \left | \vec{r}\right | = \sqrt{ 7^2+4^2+5^2 } = \sqrt{ 90 } = 3 \sqrt{10}$

The norm of a vector is also known as the length of the vector. The norm is given by the formula: 

$\displaystyle \sqrt{\sum_{i=1}^nv_i^2}=\left \| v\right \|$.

Here, we have

$\displaystyle \left \|v \right \|=\sqrt{3^2+(-12)^2+5.6^2+2^2+(-9)^2}=\sqrt{269.36}$,

the correct answer.

Write the formula to find the norm, or the length the vector.

$\displaystyle \left \| v\right \|=\sqrt{a_{1}^2+b_{2}^2}$

Substitute the known values of the vector and solve.

$\displaystyle \left \| v\right \|=\sqrt{(-2)^2+2^2}= \sqrt{4+4}=\sqrt8=2\sqrt2$

Use the following equation to find the magnitude of a vector:

$\displaystyle \left \| \vec{b}\right \|=\sqrt{i^2+j^2+k^2}$

In this case we have:

$\displaystyle \vec{b}=3i+5j-6k$

So plug in our values:

$\displaystyle \left \| \vec{b}\right \|=\sqrt{3^2+5^2+(-6)^2}=\sqrt{70}$

So:

$\displaystyle \left \| \vec{b} \right \|=\sqrt{70}$

When multiplying a vector by a scalar we multiply each component of the vector by the scalar and the result is a vector:

$\displaystyle a \vec{r}= 4 \cdot (11 \hat{i} + 7 \hat{j}) = 44 \hat{i}+28 \hat{j}$

Start by multiplying both sides by $\displaystyle r$.

$\displaystyle r=4\cos\theta$

$\displaystyle r^2=4r\cos\theta$

Keep in mind that $\displaystyle r^2=x^2+y^2$

$\displaystyle x^2+y^2=4r\cos\theta$

Remember that $\displaystyle x=r\cos\theta$

So then,

$\displaystyle x^2+y^2=4x$

$\displaystyle x^2-4x+y^2=0$

Now, complete the square.

$\displaystyle (x^2-4x+4)+y^2=4$

$\displaystyle (x-2)^2+y^2=4$

Start by multiplying both sides by $\displaystyle r$.

$\displaystyle r=8\sin\theta$

$\displaystyle r^2=8r\sin\theta$

Remember that $\displaystyle r^2=x^2+y^2$

$\displaystyle x^2+y^2=8r\sin\theta$

Keep in mind that $\displaystyle y=r\sin\theta$

So then,

$\displaystyle x^2+y^2=8y$

$\displaystyle x^2+y^2-8x=0$

Now, complete the square.

$\displaystyle x^2+(y^2-8x+16)=16$

$\displaystyle x^2+(y-4)^2=16$

This is a graph of a circle with a radius of $\displaystyle 4$ and a center at $\displaystyle (0, 4)$

Remember that $\displaystyle \sec\theta=\frac{1}{\cos\theta}$

So then $\displaystyle r=8\sec\theta$ becomes

$\displaystyle r=\frac{8}{\cos\theta}$

Now, multiply both sides by $\displaystyle \cos\theta$ to get rid of the fraction.

$\displaystyle r\cos\theta=8$

Since $\displaystyle x=r\cos\theta,$ the rectangular form of this equation is

$\displaystyle x=8$