Answer to Question #178940 in Analytic Geometry for John withers

Question

Find the cross product of (3, 2, 0) and (2, −1, −2), and use the dot product to confirm that your answer is correct. 

Answer 1

To find the cross product, we form a determinant the first row of which is a unit vector, the second row is our first vector, and the third row is our second vector

$\begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}= \begin{vmatrix} i & j & k\\ 3 & 2 & 0\\ 2 & -1 & -2 \end{vmatrix}$

$\begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}= \begin{vmatrix} 2 & 0 \\ -1 & -2 \end{vmatrix}i - \begin{vmatrix} 3 & 0 \\ 2 & -2 \end{vmatrix} j+\begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix}k$

$((2).(-2)-(0).(-1))i-((3).(-2)-(0).(2))j+((3).(-1)-(2).(2))k$

$-4i+6j-7k$

$\begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}= \begin{pmatrix} -4 \\ 6 \\ -7 \end{pmatrix}$

To confirm that this vector is perpendicular we can check the dot product is zero:

$⟨ 3, 2 , 0⟩⋅⟨ -4 , 6, -7⟩ = -12+ 12 + 0 = 0$

$⟨ 2, -1 , -2⟩⋅⟨ -4 , 6, -7⟩ = -8 - 6 + 14= 0$