Edwards Reed
01/06/2023 · Junior High School
Find the cross products \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{v} \times \mathbf{u} \) for the vectors \( \mathbf{u}=\langle 3,4,0\rangle \) and \( \mathbf{v}=\langle 0,3,-5\rangle \) \( \mathbf{u} \times \mathbf{v}=(\square) \mathbf{i}+(\square) \mathbf{j}+(\square) \mathbf{k} \) (Simplify your answers.) \( \mathbf{v} \times \mathbf{u}=(\square) \mathbf{i}+(\square) \mathbf{j}+(\square) \mathbf{k} \) (Simplify your answers.)
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Step-by-step Solution
To find the cross product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \), we can use the formula:
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} \]
Given that \( \mathbf{u} = \langle 3, 4, 0 \rangle \) and \( \mathbf{v} = \langle 0, 3, -5 \rangle \), we can substitute these values into the formula to find \( \mathbf{u} \times \mathbf{v} \).
Let's calculate \( \mathbf{u} \times \mathbf{v} \) first.
Matrices by following steps:
- step0: Solution:
\(\left|\begin{array}{cc}{i}&{j}&{k}\\{3}&{4}&{0}\\{0}&{3}&{-5}\end{array}\right|\)
- step1: Use cofactor expansion along a row:
\(-3\times \left|\begin{array}{cc}{j}&{k}\\{3}&{-5}\end{array}\right|+4\times \left|\begin{array}{cc}{i}&{k}\\{0}&{-5}\end{array}\right|-0\times \left|\begin{array}{cc}{i}&{j}\\{0}&{3}\end{array}\right|\)
- step2: Evaluate the determinant:
\(-3\left(-5j-3k\right)+4\left(-5i\right)-0\)
- step3: Evaluate:
\(-\left(-15j-9k\right)-20i-0\)
- step4: Rewrite the expression:
\(15j+9k-20i+0\)
- step5: Remove 0:
\(15j+9k-20i\)
The cross product \( \mathbf{u} \times \mathbf{v} \) is \( 15\mathbf{j} + 9\mathbf{k} - 20\mathbf{i} \).
Now, let's calculate \( \mathbf{v} \times \mathbf{u} \).
Matrices by following steps:
- step0: Solution:
\(\left|\begin{array}{cc}{i}&{j}&{k}\\{0}&{3}&{-5}\\{3}&{4}&{0}\end{array}\right|\)
- step1: Use cofactor expansion along a row:
\(-0\times \left|\begin{array}{cc}{j}&{k}\\{4}&{0}\end{array}\right|+3\times \left|\begin{array}{cc}{i}&{k}\\{3}&{0}\end{array}\right|-\left(-5\times \left|\begin{array}{cc}{i}&{j}\\{3}&{4}\end{array}\right|\right)\)
- step2: Multiply:
\(-0+3\left(-3k\right)-\left(-5\left(4i-3j\right)\right)\)
- step3: Evaluate:
\(-0-9k-\left(-20i+15j\right)\)
- step4: Rewrite the expression:
\(0-9k+20i-15j\)
- step5: Remove 0:
\(-9k+20i-15j\)
The cross product \( \mathbf{v} \times \mathbf{u} \) is \( -9\mathbf{k} + 20\mathbf{i} - 15\mathbf{j} \).
Therefore, the simplified answers are:
- \( \mathbf{u} \times \mathbf{v} = 15\mathbf{j} + 9\mathbf{k} - 20\mathbf{i} \)
- \( \mathbf{v} \times \mathbf{u} = -9\mathbf{k} + 20\mathbf{i} - 15\mathbf{j} \)
Quick Answer
- \( \mathbf{u} \times \mathbf{v} = 15\mathbf{j} + 9\mathbf{k} - 20\mathbf{i} \)
- \( \mathbf{v} \times \mathbf{u} = -9\mathbf{k} + 20\mathbf{i} - 15\mathbf{j} \)
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