Chapter 9.5, Problem 1E

The cross product of the vectors u = 〈a₁, a₂, a₃〉 and v = 〈b₁, b₂, b₃〉 is the vector

$\begin{array}{c}u\times v=|\begin{array}{ccc}i& j& k\\ \boxed{}& \boxed{}& \boxed{}\\ \boxed{}& \boxed{}& \boxed{}\end{array}|\\ =\_\_\_\_\_i-\_\_\_\_\_j+\_\_\_\_\_k\end{array}$

So the cross product of u = 〈1, 0, 1〉 and v = 〈2, 3, 0〉 is u × v = _______________.

To determine

To evaluate: The cross product of the vectors $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$, then the cross product of the vectors $u=\langle 1,0,1\rangle$ and $v=\langle 2,3,0\rangle$.

Answer to Problem 1E

The cross product of the vectors $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$ is $\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k$, then the cross product of the vectors $u=\langle 1,0,1\rangle$ and $v=\langle 2,3,0\rangle$ is $-3i+2j+3k$.

Explanation of Solution

Given:

The two vectors are $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$.

Formula used: The vectors $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$ are three-dimensional vectors, then the cross product of u and v is given below,

$u\times v=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$ (1)

Section 1:

Calculation:

Simplify the equation (1), expand the three order determinant by Row 1,

$\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|\\ =\left|\begin{array}{cc}{a}_2& a_3\\ b_2& b_3\end{array}\right|i-\left|\begin{array}{cc}{a}_1& a_3\\ b_1& b_3\end{array}\right|j+\left|\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right|k\\ =\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Thus, the cross product of the vectors $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$ is $\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k$.

Section 2:

Calculation:

The given vectors are $u=\langle 1,0,1\rangle$ and $v=\langle 2,3,0\rangle$.

Substitute 1 for $a_1$, 0 for $a_2$, 1 for $a_3$, 2 for $b_1$, 3 for $b_2$ and 0 for $b_3$ in equation (2), to find the cross product of the vectors $u=\langle 1,0,1\rangle$ and $v=\langle 2,3,0\rangle$.

$\begin{array}{c}u\times v=\left(0\times 0-1\times 3\right)i-\left(1\times 0-1\times 2\right)j+\left(1\times 3-0\times 2\right)k\\ =\left(0-3\right)i-\left(0-2\right)j+\left(3-0\right)k\\ =-3i+2j+3k\end{array}$

Thus, the cross product of the vectors $u=\langle 1,0,1\rangle$ and $v=\langle 2,3,0\rangle$ is $-3i+2j+3k$.