To find the normal vector to the plane containing vectors \(\displaystyle a\) and \(\displaystyle b\), we take the cross product of the two. 

To find the cross product between the vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), we find the determinant of the 3x3 matrix \(\displaystyle \begin{bmatrix} i&j &k \\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{bmatrix}\) which follows the formula \(\displaystyle determinant= i(a_2b_3-b_2a_3)-j(a_1b_3-b_1a_3)+k(a_1b_2-b_1a_2)\)

Applying to the vectors from the problem statement, we get

\(\displaystyle n=i(0+35)-j(0+14)+k(-5+6)=\left \langle35,-14,1\right \rangle\)

To find the normal vector to the plane containing vectors \(\displaystyle a\) and \(\displaystyle b\), we take the cross product of the two. 

To find the cross product between the vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), we find the determinant of the 3x3 matrix \(\displaystyle \begin{bmatrix} i&j &k \\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{bmatrix}\) which follows the formula \(\displaystyle determinant= i(a_2b_3-b_2a_3)-j(a_1b_3-b_1a_3)+k(a_1b_2-b_1a_2)\)

Applying to the vectors from the problem statement, we get

\(\displaystyle n=i(-9-5)-j(0-0)+k(0-0)=\left \langle-14,0,0\right \rangle\)

To find the normal vector, we must take the cross product of the two vectors.

Now, we can write the determinant in order to take the cross product of the two vectors:
\(\displaystyle \begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 8&1 &-4 \\ -1&3 &4 \end{vmatrix}\)

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

\(\displaystyle 4\hat{i}+24\hat{k}+4\hat{j}-(\hat{k}-12\hat{i}+32\hat{j})=16\hat{i}-28\hat{j}+23\hat{k}=\left \langle 16, -28, 23\right \rangle\)

The normal vector to a plane is given by the cross product of two vectors in that plane. 

So, we write the determinant in order to take the cross product of the two vectors:

\(\displaystyle \begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 10&2 &4 \\ 11&2 &6 \end{vmatrix}\)

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

\(\displaystyle 12\hat{i}+20\hat{k}+44\hat{j}-(22\hat{k}+8\hat{i}+60\hat{j})=4\hat{i}-16\hat{j}-2\hat{k}=\left \langle 4, -16, -2\right \rangle\)

To find the normal vector to a plane containing vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you take the cross product of the two vectors. 

To find the cross product between the two vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you take the determinant of the 3x3 matrix 

\(\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1& a_2&a_3 \\ b_1&b_2 &b_3 \end{bmatrix}\). \(\displaystyle determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)\)

Using the vectors from the problem statement, we get

\(\displaystyle determinant=\hat{i}(12+30)-\hat{j}(9-0)+\hat{k}(15-0)=\left \langle 42,-9,15\right \rangle\)

To find the normal vector to a plane containing vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you take the cross product of the two vectors. 

To find the cross product between the two vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you take the determinant of the 3x3 matrix 

\(\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1& a_2&a_3 \\ b_1&b_2 &b_3 \end{bmatrix}\). \(\displaystyle determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)\)

Using the vectors from the problem statement, we get

\(\displaystyle determinant=\hat{i}(-28+5)-\hat{j}(0+45)+\hat{k}(0+63)=\left \langle -23,-45,63\right \rangle\)

To find the normal vector to a plane containing vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you take the cross product of the two vectors. 

To find the cross product between the two vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you take the determinant of the 3x3 matrix 

\(\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1& a_2&a_3 \\ b_1&b_2 &b_3 \end{bmatrix}\). \(\displaystyle determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)\)

Using the vectors from the problem statement, we get

\(\displaystyle determinant=\hat{i}(-35-16)-\hat{j}(0+8)+\hat{k}(0+10)=\left \langle -51,-8,10\right \rangle\)

To obtain the normal vector to a plane containing two vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you compute the determinant of the 3x3 matrix \(\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1&a_2 &a_3 \\ b_1&b_2 & b_3 \end{bmatrix}\)

\(\displaystyle determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)\)

Using this formula, we evaluate using the vectors from the problem statement:

\(\displaystyle \hat{i}(0+18)-\hat{j}(-2-12)+\hat{k}(6-0)=\left \langle 18,14,6\right \rangle\)

To obtain the normal vector to a plane containing two vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you compute the determinant of the 3x3 matrix \(\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1&a_2 &a_3 \\ b_1&b_2 & b_3 \end{bmatrix}\)

\(\displaystyle determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)\)

Using this formula, we evaluate using the vectors from the problem statement:

\(\displaystyle \hat{i}(36-63)-\hat{j}(0-0)+\hat{k}(0-0)=\left \langle -27,0,0\right \rangle\)

To obtain the normal vector to a plane containing two vectors \(\displaystyle a=\left \langle a_1,a_2,a_3\right \rangle\) and \(\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle\), you compute the determinant of the 3x3 matrix \(\displaystyle \begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1&a_2 &a_3 \\ b_1&b_2 & b_3 \end{bmatrix}\)

\(\displaystyle determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)\)

Using this formula, we evaluate using the vectors from the problem statement:

\(\displaystyle \hat{i}(0-0)-\hat{j}(10+42)+\hat{k}(0-0)=\left \langle 0,52,0\right \rangle\)