The correct form of x,y, and z of a vector is represented in the order of i, j, and k, respectively. The coefficients of i,j, and k are used to write the vector form.

$\displaystyle -i-k = -1i+0j-1k = \left \langle -1,0,-1\right \rangle$

The x,y, and z of a vector is represented in the order of i, j, and k, respectively. Use the coefficients of i,j, and k to write the vector form.

$\displaystyle -10j=0i-10j+0k = \left \langle 0,-10,0\right \rangle$

When we are trying to find the vector form we need to remember the formula which states to take the difference between the ending and starting point.

Thus we would get:

Given $\displaystyle (a,b,c)$ and $\displaystyle (d,e,f)$ 

$\displaystyle \overrightarrow{v}=[d-a, e-b, f-c]$

In our case we have ending point at $\displaystyle (6,3,1)$ and our starting point at $\displaystyle (0,1,3)$.

Therefore we would set up the following and simplify.

$\displaystyle \overrightarrow{v}=[6-0,3-1,1-3]=[6,2,-2]$ 

The dot product will give a single value answer, and not a vector as a result.

To find the dot product, use the following formula:

$\displaystyle \left \langle x_1,x_2\right \rangle \cdot \left \langle y_1,y_2\right \rangle=(x_1)(y_1)+(x_2)(y_2)$

$\displaystyle \left \langle a,2a\right \rangle \cdot \left \langle a,a\right \rangle = (a)(a)+(2a)(a)=a^2+2a^2=3a^2$

The firing of the cannon has both x and y components.  

Write the formula that distinguishes the x and y direction and substitute. 

$\displaystyle \left \langle x,y\right \rangle= \left \langle vcos(\theta),vsin(\theta)\right \rangle$

Ensure that the calculator is in degree mode before you solve.

$\displaystyle \left \langle 25cos25,25sin25\right \rangle = \left \langle22.658,10.565 \right \rangle$

The dimensions of the vectors are mismatched.  

Since vector $\displaystyle \vec{a}$ does not have the same dimensions as $\displaystyle \vec{b}$, the answer for $\displaystyle 2\vec{a}-6\vec{b}$ cannot be solved.

To find the vector form of $\displaystyle 2i-10k$, we must map the coefficients of $\displaystyle i$, $\displaystyle j$, and $\displaystyle k$ to their corresponding $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$ coordinates. Thus, $\displaystyle 2i-10k$ becomes $\displaystyle \left \langle 2,0,-10\right \rangle$.

In order to express $\displaystyle -2i+7j-10k$ in vector form, we must use the coefficients of $\displaystyle i, j,$and $\displaystyle k$ to represent the $\displaystyle x$-, $\displaystyle y$-, and $\displaystyle z$-coordinates of the vector.

Therefore, its vector form is 

$\displaystyle \left \langle -2,7,-10\right \rangle$.

In order to express $\displaystyle i+5k$ in vector form, we must use the coefficients of $\displaystyle i, j,$and $\displaystyle k$ to represent the $\displaystyle x$-, $\displaystyle y$-, and $\displaystyle z$-coordinates of the vector.

Therefore, its vector form is 

$\displaystyle \left \langle 1,0,5\right \rangle$.

In order to express $\displaystyle -j+7k$ in vector form, we will need to map its $\displaystyle i$, $\displaystyle j$, and $\displaystyle k$ coefficients to its $\displaystyle x$-, $\displaystyle y$-, and $\displaystyle z$-coordinates.

Thus, its vector form is 

$\displaystyle \left \langle 0,-1,7\right \rangle$.