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 \(\displaystyle a=\left \langle 2,4,6\right \rangle\) , then:

\(\displaystyle \left \| a\right \|=\sqrt{(2)^{2}+(4)^{2}+(6)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{4+16+36}\)

\(\displaystyle \left \| a\right \|=\sqrt{20+36}\)

\(\displaystyle \left \| a\right \|=\sqrt{56}\)

\(\displaystyle \left \| a\right \|=\sqrt{4\times 14}\)

\(\displaystyle \left \| a\right \|=2\sqrt{14}\)

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 \(\displaystyle a=\left \langle 3,5,7\right \rangle\) , then:

\(\displaystyle \left \| a\right \|=\sqrt{(3)^{2}+(5)^{2}+(7)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{9+25+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{34+49}\)

\(\displaystyle \left \| a\right \|=\sqrt{83}\)

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 \(\displaystyle a=\left \langle9,3,6\right \rangle\) , then:

\(\displaystyle \left \| a\right \|=\sqrt{(9)^{2}+(3)^{2}+(6)^{2}}\)

\(\displaystyle \left \| a\right \|=\sqrt{81+9+36}\)

\(\displaystyle \left \| a\right \|=\sqrt{90+36}\)

\(\displaystyle \left \| a\right \|=\sqrt{126}\)

\(\displaystyle \left \| a\right \|=\sqrt{9\times 14}\)

\(\displaystyle \left \| a\right \|=3\sqrt{14}\)

In order to find the cross product of two three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if \(\displaystyle a=\left \langle a_{1},a_{2},a_{3}\right \rangle\) and \(\displaystyle b=\left \langle b_{1},b_{2},b_{3}\right \rangle\), then

\(\displaystyle a\times b=\begin{Bmatrix} i&j &k \\ a_{1}&a_{2} &a_{3} \\ b_{1}&b_{2} &b_{3} \end{Bmatrix}=\left \langle a_{2}b_{3}-a_{3}b_{2}, a_{1}b_{3}-a_{3}b_{1},a_{1}b_{2}-a_{2}b_{1} \right \rangle\).

Given \(\displaystyle a=\left \langle 5,7,2\right \rangle\) and \(\displaystyle b=\left \langle 4,4,8\right \rangle\). the cross product \(\displaystyle a \times b\) is:

\(\displaystyle \begin{Bmatrix} i&j &k \\ 5&7 &2 \\ 4&4 &8 \end{Bmatrix}\)

\(\displaystyle =\left \langle (7)(8)-(2)(4), (5)(8)-(2)(4), (5)(4)-(7)(4) \right \rangle\)

\(\displaystyle =\left \langle 56-8, 40-8, 20-28\right \rangle\)

\(\displaystyle =\left \langle 48, 32, -8\right \rangle\)

In order to find the cross product of two three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if and , then

.

Given \(\displaystyle a=\left \langle 9,8,7\right \rangle\) and \(\displaystyle b=\left \langle 0,1,2\right \rangle\), the cross product is:

\(\displaystyle \begin{Bmatrix} i&j &k \\ 9&8 &7 \\ 0&1 &2 \end{Bmatrix}\)

\(\displaystyle =\left \langle (8)(2)-(7)(1), (9)(2)-(7)(0), (9)(1)-(8)(0) \right \rangle\)

\(\displaystyle =\left \langle 16-7, 18-0, 9-0 \right \rangle\)

\(\displaystyle =\left \langle 9, 18, 9 \right \rangle\)

In order to find the cross product of two three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if  and , then

.

Given \(\displaystyle a=\left \langle 7,-4,3\right \rangle\) and \(\displaystyle b=\left \langle 5,-5,5\right \rangle\), the cross product  is:

\(\displaystyle \begin{Bmatrix} i&j &k \\ 7&-4 &3 \\ 5&-5 &5 \end{Bmatrix}\)

\(\displaystyle =\left \langle (-4)(5)-(3)(-5), (7)(5)-(3)(5), (7)(-5)-(-4)(5) \right \rangle\)

\(\displaystyle =\left \langle -20+15, 35-15, -35+20 \right \rangle\)

\(\displaystyle =\left \langle -5, 20, -15 \right \rangle\)

In order to find the cross product of two three-dimensional vectors, we must find the determinant of a matrix comprised of the vectors' elements. That is, if  and , then

.

Given \(\displaystyle a=\left \langle 0,-3,3\right \rangle\) and \(\displaystyle b=\left \langle -3,0,3\right \rangle\), the cross product  is:

\(\displaystyle \begin{Bmatrix} i&j &k \\ 0&-3 &3 \\ -3&0 &3 \end{Bmatrix}\)

\(\displaystyle =\left \langle (-3)(3)-(3)(0), (0)(3)-(3)(-3), (0)(0)-(-3)(-3) \right \rangle\)

\(\displaystyle =\left \langle -9-0, 0+9, 0-9 \right \rangle\)

\(\displaystyle =\left \langle -9, 9,-9 \right \rangle\)

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle 3,5,7\right \rangle\) and \(\displaystyle b=\left \langle -4,-1,2\right \rangle\), then:

\(\displaystyle \left \langle 3,5,7\right \rangle\times \left \langle -4,-1,2\right \rangle\)

\(\displaystyle =(3\times -4)+(5\times -1)+(7\times 2)\)

\(\displaystyle =(-12)+(-5)+(14)\)

\(\displaystyle =-17+14\)

\(\displaystyle =-3\)

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle 2,0,-2\right \rangle\) and \(\displaystyle b=\left \langle 6,1,7\right \rangle\) then:

\(\displaystyle \left \langle 2,0,-2\right \rangle\times \left \langle 6,1,7\right \rangle\)

\(\displaystyle =(2\times 6)+(0\times 1)+(-2\times7)\)

\(\displaystyle =12+0-14\)

\(\displaystyle =-2\)

The dot product of two vectors is the sum of the products of the vectors' corresponding elements. Given \(\displaystyle a=\left \langle 0,5,1\right \rangle\) and \(\displaystyle b=\left \langle 5,1,0\right \rangle\), then:

\(\displaystyle \left \langle 0,5,1\right \rangle\times\left \langle 5,1,0\right \rangle\)

\(\displaystyle =(0\times5)+(5\times1)+(1\times0)\)

\(\displaystyle =0+5+0\)

\(\displaystyle =5\)