Let u and v be vectors in ℝ3 and let c be a scalar. Prove that c(u×v) =(cu)×v =u×(cv). (This is Theorem 10.28).

Hence, we prove that c(u×v) =(cu)×v =u×(cv).

Q. 67 - Calculus | StudyQuestionHub

Q. 67

Question

Let u and v be vectors in $ℝ^{3}$ and let c be a scalar. Prove that $c (u \times v) = (c u) \times v = u \times (c v)$ . (This is Theorem 10.28).

Step-by-Step Solution

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Answer

Hence, we prove that $c (u \times v) = (c u) \times v = u \times (c v)$ .

1Step 1. Given Information

Let u and v be vectors in $ℝ^{3}$ and let c be a scalar. Prove that $c (u \times v) = (c u) \times v = u \times (c v)$ .

2Step 2. We have to prove c ( u × v )   = ( cu ) × v   = u × ( cv )

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

Firstly finding the value of $c (u \times v)$

$c (u \times v) = c {(u_{1}, u_{2}, u_{3}) \times (v_{1}, v_{2}, v_{3})} c (u \times v) = c [\begin{matrix} i & j & k \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix}] c (u \times v) = c (i |\begin{matrix} u_{2} & u_{3} \\ v_{2} & v_{3} \end{matrix}| - j |\begin{matrix} u_{1} & u_{3} \\ v_{1} & v_{3} \end{matrix}| + k |\begin{matrix} u_{1} & u_{2} \\ v_{1} & v_{2} \end{matrix}|) c (u \times v) = c [i (u_{2} v_{3} - u_{3} v_{2}) - j (u_{1} v_{3} - u_{3} v_{1}) + k (u_{1} v_{2} - u_{2} v_{1})] c (u \times v) = c (u_{2} v_{3} - u_{3} v_{2}, - u_{1} v_{3} + u_{3} v_{1}, u_{1} v_{2} - u_{2} v_{1}) c (u \times v) = (c u_{2} v_{3} - c u_{3} v_{2}, - c u_{1} v_{3} + c u_{3} v_{1}, c u_{1} v_{2} - c u_{2} v_{1})$

3Step 3. Now finding the value of ( c u ) × v

$(c u) \times v = (c u_{1}, c u_{2}, c u_{3}) \times (v_{1}, v_{2}, v_{3}) (c u) \times v = [\begin{matrix} i & j & k \\ c u_{1} & c u_{2} & c u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix}] (c u) \times v = i |\begin{matrix} c u_{2} & c u_{3} \\ v_{2} & v_{3} \end{matrix}| - j |\begin{matrix} c u_{1} & c u_{3} \\ v_{1} & v_{3} \end{matrix}| + k |\begin{matrix} c u_{1} & c u_{2} \\ v_{1} & v_{2} \end{matrix}| (c u) \times v = i (c u_{2} v_{3} - c u_{3} v_{2}) - j (c u_{1} v_{3} - c u_{3} v_{1}) + k (c u_{1} v_{2} - c u_{2} v_{1}) (c u) \times v = (c u_{2} v_{3} - c u_{3} v_{2}, - c u_{1} v_{3} + c u_{3} v_{1}, c u_{1} v_{2} - c u_{2} v_{1})$

4Step 4. Now finding the value of u × ( c v )

$u \times (c v) = (u_{1}, u_{2}, u_{3}) \times (c v_{1}, c v_{2}, c v_{3}) u \times (c v) = [\begin{matrix} i & j & k \\ u_{1} & u_{2} & u_{3} \\ c v_{1} & c v_{2} & c v_{3} \end{matrix}] u \times (c v) = i |\begin{matrix} u_{2} & u_{3} \\ c v_{2} & c v_{3} \end{matrix}| - j |\begin{matrix} u_{1} & u_{3} \\ c v_{1} & c v_{3} \end{matrix}| + k |\begin{matrix} u_{1} & u_{2} \\ c v_{1} & c v_{2} \end{matrix}| u \times (c v) = i (c u_{2} v_{3} - c u_{3} v_{2}) - j (c u_{1} v_{3} - c u_{3} v_{1}) + k (c u_{1} v_{2} - c u_{2} v_{1}) u \times (c v) = (c u_{2} v_{3} - c u_{3} v_{2}, - c u_{1} v_{3} + c u_{3} v_{1}, c u_{1} v_{2} - c u_{2} v_{1})$

Hence, prove that $c (u \times v) = (cu) \times v = u \times (cv)$

Q. 66

Q. 68

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