Let vector \(\displaystyle \vec{a}=\left \langle a1,a2\right \rangle\)  and \(\displaystyle \vec{b}=\left \langle b1,b2\right \rangle\).

The dot product is equal to:

\(\displaystyle \vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2\)

Following this rule for the current problem, simplify.

\(\displaystyle \left \langle 3,2\right \rangle\cdot\left \langle -6,9\right \rangle = (3)(-6)+(2)(9)=0\)

Let vector \(\displaystyle \vec{a}=\left \langle a1,a2,a3\right \rangle\)  and \(\displaystyle \vec{b}=\left \langle b1,b2,b3\right \rangle\).

Use the following formula to solve this dot product:

 \(\displaystyle \vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2+a3\cdot b3\)

Substitute and solve.

\(\displaystyle \left \langle3,-1,2 \right \rangle \cdot \left \langle -2,4,5\right \rangle=(3)(-2)+(-1)(4)+(2)(5 )=0\)

The vectors can be rewritten in the following form:

\(\displaystyle \begin{vmatrix} i& j& k\\ 1&-1 &0 \\ 0& 0& -1 \end{vmatrix}\)

One method of solving the 3 by 3 determinant is to break this down into 2 by 2 determinants. The determinant of 2 by 2 matrices are computed by \(\displaystyle ad-bc\), such that:

\(\displaystyle D=\begin{vmatrix} a&b \\ c& d \end{vmatrix}\)

Rewrite the 3 by 3 determinant.

\(\displaystyle =i\begin{vmatrix} -1& 0\\ 0&-1 \end{vmatrix}- j\begin{vmatrix} 1&0 \\ 0&-1 \end{vmatrix}+k\begin{bmatrix} 1&-1 \\ 0&0 \end{bmatrix}\)

\(\displaystyle =i[(-1)(-1)-0] -j[(1)(-1)-0]+k[0]\)

\(\displaystyle = i[1]-j[-1]\)

\(\displaystyle = i+j\)

The horizontal and vertical components are shown below:

 \(\displaystyle Vx=Vcos\Theta\)

\(\displaystyle Vy=Vsin\Theta\)

Plug in the velocity and the given angle to the equations.

\(\displaystyle Vx=Vcos\Theta=10cos(10) = 9.848\)

\(\displaystyle Vy=Vsin\Theta = 10sin(10)=1.736\)

Therefore, the components in vector form is \(\displaystyle \left \langle 9.848,1.736\right \rangle\).

Let vectors \(\displaystyle \vec{a}=\left \langle a1,a2\right \rangle\)  and \(\displaystyle \vec{b}=\left \langle b1,b2\right \rangle\).

The formula for the dot product is:

\(\displaystyle \vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2\)

Follow this formula and simplify.

\(\displaystyle \left \langle 1,6\right \rangle\cdot\left \langle 1,6\right \rangle = (1)(1)+(6)(6)=37\)

The problem \(\displaystyle \left \langle 3,4,5\right \rangle\cdot\left \langle1,2,3 \right \rangle\) is in the form of a dot product.  The final answer must be an integer, and not in vector form.

Write the formula for the dot product.

\(\displaystyle \left \langle a_1,a_2,a_3\right \rangle\cdot\left \langle b_1,b_2,b_3\right \rangle=a_1b_1+a_2b_2+a_3b_3\)

Substitute the givens and solve.

\(\displaystyle (3)(1)+(4)(2)+(5)(3)=3+8+15=26\)

Calculate \(\displaystyle 2\vec{a}\).

\(\displaystyle \vec{a}= \left \langle 2,4,-3\right \rangle\)

\(\displaystyle 2\vec{a}= \left \langle 4,8,-6\right \rangle\)

Find the magnitude.

\(\displaystyle |2\vec{a}|=\sqrt{4^2+8^2+(-6)^2}=\sqrt{16+64+36}=\sqrt{116}=2\sqrt{29}\)

In order for the particles to collide, their \(\displaystyle \small x\) and \(\displaystyle \small y\) coordinates must be equal simultaneously. In order to check if this happens, we can set the particles' \(\displaystyle \small x\)-coordinates and \(\displaystyle \small y\)-coordinates equal to each other.

Let's start with the \(\displaystyle \small x\)-coordinate: 

\(\displaystyle \small \small \small 3t^{2}-t=6\rightarrow 3t^2-t-6=0\).

Using the quadratic formula

\(\displaystyle \small \small t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), we get \(\displaystyle \small \small t=\frac{1+\sqrt{73}}{6}\approx1.59\) 

the other root is negative, so it can be discarded since \(\displaystyle \small t>0\).

Now let's do the \(\displaystyle \small y\)-coordinate: \(\displaystyle \small \small t+1/t=4\rightarrow t^2+1=4t\rightarrow t^2-4t+1=0\). Use the quadratic formula again to solve for \(\displaystyle \small t\), and you'll get \(\displaystyle \small 2\pm\sqrt{3}\) (these roots are approximately \(\displaystyle \small 0.27\) and \(\displaystyle \small 3.73\)). The particles never have the same \(\displaystyle \small x\) and \(\displaystyle \small y\) coordinate simultaneously, so they do not collide.

\(\displaystyle \vec{a}=< 3,2,1>\)

\(\displaystyle \vec{b}=< 1,0,10>\)

\(\displaystyle \vec{a}\cdot\vec{b}\) is simply the dot product of these two vectors. Mathematically, this is calculated as follows.

\(\displaystyle (3\cdot 1)+(2\cdot 0)+(1\cdot 10)\)  \(\displaystyle =13\)

To find the dot product of \(\displaystyle a=\left \langle 3,4,7\right \rangle\) and \(\displaystyle b=\left \langle -1,5,6\right \rangle\), calculate the sum of the products of the vectors' corresponding components:

\(\displaystyle \left \langle 3,4,7\right \rangle\times \left \langle -1,5,6\right \rangle\)

\(\displaystyle =(3\times -1)+ (4\times 5)+(7\times 6)\)

\(\displaystyle =-3+ 20+42\)

\(\displaystyle =-3+ 62\)

\(\displaystyle =59\)