Chapter 10: Q. 69 (page 825)

Let u and v be vectors in $ℝ^{3}$ . Prove that $v \cdot (u \times v) = 0$ . (This is Theorem 10.31(b).)

Short Answer

Expert verified

Hence, we prove that $v \cdot (u \times v) = 0$

Step by step solution

Step 1. Given Information

Let u and v be vectors in $ℝ^{3}$ . Prove that role="math" localid="1649924080419" $v \cdot (u \times v) = 0$ .

Step 2. We have to prove v·(u×v)=0

Let $u = (u_{1}, u_{2}, u_{3}) and v = (v_{1}, v_{2}, v_{3})$

Now the determinant is

$v \cdot (u \times v) = \det [\begin{matrix} v_{1} & v_{2} & v_{3} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix}] v \cdot (u \times v) = v_{1} |\begin{matrix} u_{2} & u_{3} \\ v_{2} & v_{3} \end{matrix}| - v_{2} |\begin{matrix} u_{1} & u_{3} \\ v_{1} & v_{3} \end{matrix}| + v_{3} |\begin{matrix} u_{1} & u_{2} \\ v_{1} & v_{2} \end{matrix}| v \cdot (u \times v) = v_{1} (u_{2} v_{3} - u_{3} v_{2}) - v_{2} (u_{1} v_{3} - u_{3} v_{1}) + v_{3} (u_{1} v_{2} - u_{2} v_{1}) v \cdot (u \times v) = v_{1} u_{2} v_{3} - v_{1} u_{3} v_{2} - v_{2} u_{1} v_{3} + v_{2} u_{3} v_{1} + v_{3} u_{1} v_{2} - v_{3} u_{2} v_{1} v \cdot (u \times v) = 0$

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Let u and v be vectors in $ℝ^{3}$ . Prove that $v \cdot (u \times v) = 0$ . (This is Theorem 10.31(b).)

Short Answer

Step by step solution

Step 1. Given Information

Step 2. We have to prove v·(u×v)=0

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Let u and v be vectors in ℝ3. Prove that v·(u×v)=0. (This is Theorem 10.31(b).)

Short Answer

Step by step solution

Step 1. Given Information

Step 2. We have to prove v·(u×v)=0

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Discrete Mathematics

Mechanics Maths

Pure Maths

Applied Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Let u and v be vectors in $ℝ^{3}$ . Prove that $v \cdot (u \times v) = 0$ . (This is Theorem 10.31(b).)