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벡터의 연산

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

CLP-3 · 1.2절 · 37번대학 수준우리말 번역

a×(b×c)=(a⋅c)b−(a⋅b)c\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) =(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c} 임을 보이세요.

영어 원문 보기

Show that a×(b×c)=(a⋅c)b−(a⋅b)c\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) =(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}.

정답·풀이 보기

정의에 의해,

b×c = (b2c3−b3c2)ı^−(b1c3−b3c1)ȷ^+(b1c2−b2c1)k^따라서 좌변과 우변은 다음과 같다a×(b×c) = det⁡[ı^ȷ^k^a1a2a3b2c3−b3c2−b1c3+b3c1b1c2−b2c1] = ı^ [a2(b1c2−b2c1)−a3(−b1c3+b3c1)]−ȷ^ [a1(b1c2−b2c1)−a3(b2c3−b3c2)]+k^ [a1(−b1c3+b3c1)−a2(b2c3−b3c2)](lhs)(a⋅c)b−(a⋅b)c = (a1c1+a2c2+a3c3)(b1ı^+b2ȷ^+b3k^)−(a1b1+a2b2+a3b3)(c1ı^+c2ȷ^+c3k^)= ı^ [a1b1c1+a2b1c2+a3b1c3−a1b1c1−a2b2c1−a3b3c1]+ȷ^ [a1b2c1+a2b2c2+a3b2c3−a1b1c2−a2b2c2−a3b3c2]+k^ [a1b3c1+a2b3c2+a3b3c3−a1b1c3−a2b2c3−a3b3c3]= ı^ [a2b1c2+a3b1c3−a2b2c1−a3b3c1]+ȷ^ [a1b2c1+a3b2c3−a1b1c2−a3b3c2]+k^ [a1b3c1+a2b3c2−a1b1c3−a2b2c3](rhs)\begin{aligned} \mathbf{b}\times\mathbf{c} \ =\ &(b_2c_3-b_3c_2)\hat{\boldsymbol{\imath}}-(b_1c_3-b_3c_1)\hat{\boldsymbol{\jmath}} +(b_1c_2-b_2c_1)\hat{\mathbf{k}} \\ \\ \text{따라서 좌변과 우변은 다음과 같다} \\ \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) \ =\ &\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ a_1&a_2&a_3\\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1\end{matrix}\right]\\ \ =\ &\hat{\boldsymbol{\imath}}\,[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)]\\ {-}&\hat{\boldsymbol{\jmath}}\,[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\\ {+}&\hat{\mathbf{k}}\,[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)]&{\rm (lhs)}\\ (\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c} \ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hat{\boldsymbol{\imath}}+b_2\hat{\boldsymbol{\jmath}}+b_3\hat{\mathbf{k}}) -(a_1b_1+a_2b_2+a_3b_3)(c_1\hat{\boldsymbol{\imath}}+c_2\hat{\boldsymbol{\jmath}}+c_3\hat{\mathbf{k}})\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_1b_1c_2-a_2b_2c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_1b_1c_3-a_2b_2c_3-a_3b_3c_3]\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] &{\rm (rhs)} \end{aligned}

(lhs)와 (rhs)는 같다.

원본 교재 풀이(영어)

By definition,

b×c = (b2c3−b3c2)ı^−(b1c3−b3c1)ȷ^+(b1c2−b2c1)k^so that the left and right hand sides area×(b×c) = det⁡[ı^ȷ^k^a1a2a3b2c3−b3c2−b1c3+b3c1b1c2−b2c1] = ı^ [a2(b1c2−b2c1)−a3(−b1c3+b3c1)]−ȷ^ [a1(b1c2−b2c1)−a3(b2c3−b3c2)]+k^ [a1(−b1c3+b3c1)−a2(b2c3−b3c2)](lhs)(a⋅c)b−(a⋅b)c = (a1c1+a2c2+a3c3)(b1ı^+b2ȷ^+b3k^)−(a1b1+a2b2+a3b3)(c1ı^+c2ȷ^+c3k^)= ı^ [a1b1c1+a2b1c2+a3b1c3−a1b1c1−a2b2c1−a3b3c1]+ȷ^ [a1b2c1+a2b2c2+a3b2c3−a1b1c2−a2b2c2−a3b3c2]+k^ [a1b3c1+a2b3c2+a3b3c3−a1b1c3−a2b2c3−a3b3c3]= ı^ [a2b1c2+a3b1c3−a2b2c1−a3b3c1]+ȷ^ [a1b2c1+a3b2c3−a1b1c2−a3b3c2]+k^ [a1b3c1+a2b3c2−a1b1c3−a2b2c3](rhs)\begin{aligned} \mathbf{b}\times\mathbf{c} \ =\ &(b_2c_3-b_3c_2)\hat{\boldsymbol{\imath}}-(b_1c_3-b_3c_1)\hat{\boldsymbol{\jmath}} +(b_1c_2-b_2c_1)\hat{\mathbf{k}} \\ \\ \text{so that the left and right hand sides are} \\ \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) \ =\ &\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ a_1&a_2&a_3\\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1\end{matrix}\right]\\ \ =\ &\hat{\boldsymbol{\imath}}\,[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)]\\ {-}&\hat{\boldsymbol{\jmath}}\,[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\\ {+}&\hat{\mathbf{k}}\,[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)]&{\rm (lhs)}\\ (\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c} \ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hat{\boldsymbol{\imath}}+b_2\hat{\boldsymbol{\jmath}}+b_3\hat{\mathbf{k}}) -(a_1b_1+a_2b_2+a_3b_3)(c_1\hat{\boldsymbol{\imath}}+c_2\hat{\boldsymbol{\jmath}}+c_3\hat{\mathbf{k}})\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_1b_1c_2-a_2b_2c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_1b_1c_3-a_2b_2c_3-a_3b_3c_3]\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] &{\rm (rhs)} \end{aligned}

(lhs) and (rhs) are the same.

한 번 더 — 비슷한 문제OpenStax 3권 · 2장 복습 · 13번우리말 번역

다음 문제에서 주어진 조건을 만족시키는 직선의 벡터방정식과 매개변수방정식을 구하세요.

두 점 (1,3,5)(1,3,5) 와 (−2,6,−3)(-2,6,-3) 을 지나는 직선

다 풀었으면 정답 확인

x=1−3t,y=3+3t,z=5−8t,r(t)=(1−3t)i+3(1+t)j+(5−8t)kx=1-3t,y=3+3t,z=5-8t,\text{r}(t)=(1-3t)\text{i}+3(1+t)\text{j}+(5-8t)\text{k}

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