The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle 3,-3,-3\right \rangle\) and \(\displaystyle b=\left \langle -3,3,-3\right \rangle\):

\(\displaystyle \left \langle 3,-3,-3\right \rangle\times \left \langle -3,3,-3\right \rangle\)

\(\displaystyle =(3\times -3)+(-3\times 3)+(-3\times -3)\)

\(\displaystyle =(-9)+(-9)+(9)\)

\(\displaystyle =-18+9\)

\(\displaystyle =-9\)  

The dot product of two vectors is the sum of the products of the vectors' composite elements. Thus, given \(\displaystyle a=\left \langle -1,-1,1\right \rangle\) and \(\displaystyle b=\left \langle 1,-1,-1\right \rangle\).

\(\displaystyle \left \langle -1,-1,1\right \rangle\times\left \langle 1,-1,-1\right \rangle\)

\(\displaystyle =(-1\times 1)+(-1\times -1)+(1\times -1)\)

\(\displaystyle =(-1)+(1)+(-1)\)

\(\displaystyle =0+(-1)\)

\(\displaystyle =-1\)

The dot product of two vectors is the sum of the products of the vectors' composite elements. Thus, given \(\displaystyle a=\left \langle 2,9,7\right \rangle\) and \(\displaystyle b=\left \langle 1,3,6 \right \rangle\).

\(\displaystyle \left \langle 2,9,7\right \rangle\times \left \langle1,3,6 \right \rangle\)

\(\displaystyle =(2\times 1)+(9\times 3)+(7\times 6)\)

\(\displaystyle =2+27+42\)

\(\displaystyle =2+69\)

\(\displaystyle =71\)

To evaluate the dot product, apply the following formula:

\(\displaystyle \left \langle x_1,y_1 \right \rangle \cdot \left \langle x_2,y_2 \right \rangle= x_1x_2+ y_1y_2\)

\(\displaystyle \left \langle a^2,2a^2\right \rangle \cdot \left \langle 2a^2,a^2\right \rangle= (a^2)(2a^2)+(2a^2)(a^2)= 2a^4+2a^4\)

\(\displaystyle \left \langle a^2,2a^2\right \rangle \cdot \left \langle 2a^2,a^2\right \rangle=4a^4\)

Do not mistaken the times symbol for multiplication.  This is a notation for computing the cross product of two vectors.

Write the formula to compute the cross product of two vectors.

For \(\displaystyle \vec{x}= \left \langle x_1,x_2,x_3\right \rangle\) and \(\displaystyle \vec{y}= \left \langle y_1,y_2,y_3\right \rangle\):

\(\displaystyle \vec{x}\times \vec{y}=\left \langle x_2y_3-x_3y_2, x_3y_1-x_1y_3,x_1y_2-x_2y_1 \right \rangle\)

Substitute the values and solve for the cross product.

\(\displaystyle \vec{x}\times \vec{y}=\left \langle (2)(6)-(3)(5), (3)(4)-(1)(6), (1)(5)-(2)(4) \right \rangle\)

\(\displaystyle \vec{x}\times \vec{y}=\left \langle -3,6,-3\right \rangle\)

The dot product of two vectors is the sum of the products of the vectors' composite elements. Thus, given \(\displaystyle a=\left \langle 3,7,-10\right \rangle\) and \(\displaystyle b=\left \langle 4,6,10 \right \rangle\).

\(\displaystyle \left \langle 3,7,-10\right \rangle\times \left \langle 4,6,10 \right \rangle\) 

\(\displaystyle =(3\times 4)+(7\times 6)+(-10\times 10)\)

\(\displaystyle =12+42+(-100)\)

\(\displaystyle =54+(-100)\)

\(\displaystyle =-46\)

In order to find the norm of a vector, we must take the square root of the sums of the squares of the vector's elements. Given \(\displaystyle a=\left \langle -2,-1,1\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(-2)^{2}+(-1)^{2}+(1)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{4+1+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{6}\)

In order to find the norm of a vector, we must take the square root of the sums of the squares of the vector's elements. Given \(\displaystyle a=\left \langle 4,0,5\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(4)^{2}+(0)^{2}+(5)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{16+0+25}\)

\(\displaystyle \left \| a\right \|=\sqrt{41}\)

In order to find the norm of a vector, we must take the square root of the sums of the squares of the vector's elements. Given \(\displaystyle a=\left \langle 1,0,-1\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(1)^{2}+(0)^{2}+(-1)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{1+0+1}\)

\(\displaystyle \left \| a\right \|=\sqrt{2}\)

In order to find the norm of a vector, we must first find the sum of the squares of the vector's elements and take the square root of that sum. Given \(\displaystyle a=\left \langle 5,-6,7\right \rangle\), then:

\(\displaystyle \left \| a\right \|=\sqrt{(5)^{2}+(-6)^{2}+(7)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{25+36+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{61+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{110}\)