Pertanyaan

Tentukan dua vektor satuan yang masing-masing tegak lurus pada vektor dan . Tentukan juga sudut antara masing-masing vektor satuan ini dengan vektor .

Tentukan dua vektor satuan yang masing-masing tegak lurus pada vektor $\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $3\overset{\boldsymbol\hat{}}{\boldsymbol i}+2\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ .
Tentukan juga sudut antara masing-masing vektor satuan ini dengan vektor $\frac12\left(\overset{\boldsymbol\hat{}}{\boldsymbol i}-\overset{\mathit\hat{}}{\boldsymbol j}-\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)$ .

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Jawaban

sudut antara masing-masing vektor satuan ini dengan vektor adalah θ 1 = 54 , 7 4 ∘ dan θ 2 = 125 , 2 6 ∘ .

sudut antara masing-masing vektor satuan ini dengan vektor $\frac12\left(\overset{\boldsymbol\hat{}}{\boldsymbol i}-\overset{\mathit\hat{}}{\boldsymbol j}-\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)$ adalah $θ_{1} = 54, 7 4^{\circ}$ dan $θ_{2} = 125, 2 6^{\circ}$ .

Pembahasan

Jawaban yang benar untuk bagian a adalah dan . Jawaban yang benar untuk bagian b adalah θ 1 = 54 , 7 4 ∘ dan θ 2 = 125 , 2 6 ∘ . Bagian a: Ingat bahwa hasil perkalian dot product antara dua vektor dan yang saling tegak lurus adalah 0. Sehingga, u ⋅ v u 1 v 1 + u 2 v 2 + u 3 v 3 = = 0 0 Panjang vektor tiga dimensi dapat dihitung menggunakan rumus berikut: ∣ u ∣ = u 1 2 + u 2 2 + u 3 2 Misalkan vektor satuan yang tegak lurus terhadapvektor dan adalahvektor . Maka: dan Substitusi b = − 4 c ke persamaan 3 a + b + 4 c = 0 : 3 a + 2 b + 4 c 3 a + 2 ( − 4 c ) + 4 c 3 a − 8 c + 4 c 3 a − 4 c 3 a a = = = = = = 0 0 0 0 4 c 3 4 c Karena vektor merupakan vektor satuan, maka panjang vektornya adalah 1. Sehingga, Substitusi b = − 4 c dan a = 3 4 c ke persamaan a 2 + b 2 + c 2 = 1 : a 2 + b 2 + c 2 ( 3 4 c ) 2 + ( − 4 c ) 2 + c 2 9 16 c 2 + 16 c 2 + c 2 9 16 c 2 + 144 c 2 + 9 c 2 169 c 2 c 2 c = = = = = = = = 1 1 1 1 9 169 9 ± 169 9 ± 13 3 Dari perhitungan di atas, diperoleh c 1 = 13 3 dan c 2 = − 13 3 . Untuk c 1 = 13 3 , maka: b 1 a 1 = = = = = = − 4 c 1 − 4 ( 13 3 ) − 13 12 3 4 c 1 3 4 ( 13 3 ) 13 4 Untuk c 2 = − 13 3 , maka: b 2 a 1 = = = = = = − 4 c 2 − 4 ( − 13 3 ) 13 12 3 4 c 1 3 4 ( − 13 3 ) − 13 4 Sehingga vektor yang tegak lurus terhadap vektor dan , yaitu: Dengan demikian, dua vektor yang tegak lurus terhadapvektor dan adalah dan . Bagian b: Ingat bahwa sudut antara dua vektor dan dapat dihitung menggunakan rumus berikut: cos θ = = ∣ u ∣ ∣ v ∣ u ⋅ v ( u 1 2 + u 2 2 + u 3 2 ) ( v 1 2 + v 2 2 + v 3 2 ) u 1 v 1 + u 2 v 2 + u 3 v 3 Sudut antara dan , yaitu: Sudut antara dan , yaitu: Dengan demikian,sudut antara masing-masing vektor satuan ini dengan vektor adalah θ 1 = 54 , 7 4 ∘ dan θ 2 = 125 , 2 6 ∘ .

Jawaban yang benar untuk bagian a adalah ${\boldsymbol v}_{\mathbf1}=\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}+\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan ${\boldsymbol v}_{\mathbf2}=-\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}+\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}-\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}$ .

Jawaban yang benar untuk bagian b adalah $θ_{1} = 54, 7 4^{\circ}$ dan $θ_{2} = 125, 2 6^{\circ}$ .

Bagian a:

Ingat bahwa hasil perkalian dot product antara dua vektor $\boldsymbol u=u_1\overset{\boldsymbol\hat{}}{\boldsymbol i}+u_2\overset{\boldsymbol\hat{}}{\boldsymbol j}+u_3\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $\boldsymbol v=v_1\overset{\boldsymbol\hat{}}{\boldsymbol i}+v_2\overset{\boldsymbol\hat{}}{\boldsymbol j}+v_3\overset{\boldsymbol\hat{}}{\boldsymbol k}$ yang saling tegak lurus adalah 0. Sehingga,

$u \cdot v u_{1} v_{1} + u_{2} v_{2} + u_{3} v_{3} = = 00$

Panjang vektor tiga dimensi $\boldsymbol u=u_1\overset{\boldsymbol\hat{}}{\boldsymbol i}+u_2\overset{\boldsymbol\hat{}}{\boldsymbol j}+u_3\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dapat dihitung menggunakan rumus berikut:

$∣ u ∣ = u_{1}^{2} + u_{2}^{2} + u_{3}^{2}$

Misalkan vektor satuan yang tegak lurus terhadap vektor $\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $3\overset{\boldsymbol\hat{}}{\boldsymbol i}+2\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ adalah vektor $a\overset{\boldsymbol\hat{}}{\boldsymbol i}+b\overset{\mathit\hat{}}{\boldsymbol j}+c\overset{\boldsymbol\hat{}}{\boldsymbol k}$ . Maka:

$\begin{array}{rcl}\left(a\overset{\boldsymbol\hat{}}{\boldsymbol i}+b\overset{\mathit\hat{}}{\boldsymbol j}+c\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)\cdot\left(\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)&=&0\\\left(a\right)\left(0\right)+\left(b\right)\left(1\right)+\left(c\right)\left(4\right)&=&0\\b+4c&=&0\\b&=&-4c\end{array}$

dan

$\begin{array}{rcl}\left(a\overset{\boldsymbol\hat{}}{\boldsymbol i}+b\overset{\mathit\hat{}}{\boldsymbol j}+c\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)\cdot\left(3\overset{\boldsymbol\hat{}}{\boldsymbol i}+2\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)&=&0\\\left(a\right)\left(3\right)+\left(b\right)\left(2\right)+\left(c\right)\left(4\right)&=&0\\3a+2b+4c&=&0\end{array}$

Substitusi $b = - 4 c$ ke persamaan $3 a + b + 4 c = 0$ :

$3 a + 2 b + 4 c 3 a + 2 (- 4 c) + 4 c 3 a - 8 c + 4 c 3 a - 4 c 3 a a = = = = = = 0000 4 c \frac{4}{3} c$

Karena vektor $a\overset{\boldsymbol\hat{}}{\boldsymbol i}+b\overset{\mathit\hat{}}{\boldsymbol j}+c\overset{\boldsymbol\hat{}}{\boldsymbol k}$ merupakan vektor satuan, maka panjang vektornya adalah 1. Sehingga,

$\begin{array}{rcl}\left|a\overset{\boldsymbol\hat{}}{\mathbf i}+b\overset{\boldsymbol\hat{}}{\mathbf j}+c\overset{\boldsymbol\hat{}}{\mathbf k}\right|&=&1\\\sqrt{a^2+b^2+c^2}&=&1\\a^2+b^2+c^2&=&1\end{array}$

Substitusi $b = - 4 c$ dan $a = \frac{4}{3} c$ ke persamaan $a^{2} + b^{2} + c^{2} = 1$ :

$a^{2} + b^{2} + c^{2} (\frac{4}{3} c)^{2} + (- 4 c)^{2} + c^{2} \frac{16}{9} c^{2} + 16 c^{2} + c^{2} \frac{16 c ^{2} + 144 c ^{2} + 9 c ^{2}}{9} 169 c^{2} c^{2} c = = = = = = = = 11119 \frac{9}{169} \pm \frac{9}{169} \pm \frac{3}{13}$

Dari perhitungan di atas, diperoleh $c_{1} = \frac{3}{13}$ dan $c_{2} = - \frac{3}{13}$ . Untuk $c_{1} = \frac{3}{13}$ , maka:

$b_{1} a_{1} = = = = = = - 4 c_{1} - 4 (\frac{3}{13}) - \frac{12}{13} \frac{4}{3} c_{1} \frac{4}{3} (\frac{3}{13}) \frac{4}{13}$

Untuk $c_{2} = - \frac{3}{13}$ , maka:

$b_{2} a_{1} = = = = = = - 4 c_{2} - 4 (- \frac{3}{13}) \frac{12}{13} \frac{4}{3} c_{1} \frac{4}{3} (- \frac{3}{13}) - \frac{4}{13}$

Sehingga vektor yang tegak lurus terhadap vektor $\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $3\overset{\boldsymbol\hat{}}{\boldsymbol i}+2\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ , yaitu:

$\begin{array}{rcl}{\boldsymbol v}_{\mathbf1}&=&\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}+\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}\\&&\\{\boldsymbol v}_{\mathbf2}&=&-\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}+\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}-\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}\end{array}$

Dengan demikian, dua vektor yang tegak lurus terhadap vektor $\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $3\overset{\boldsymbol\hat{}}{\boldsymbol i}+2\overset{\mathit\hat{}}{\boldsymbol j}+4\overset{\boldsymbol\hat{}}{\boldsymbol k}$ adalah ${\boldsymbol v}_{\mathbf1}=\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}+\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan ${\boldsymbol v}_{\mathbf2}=-\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}+\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}-\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}$ .

Bagian b:

Ingat bahwa sudut antara dua vektor $\boldsymbol u=u_1\overset{\boldsymbol\hat{}}{\boldsymbol i}+u_2\overset{\boldsymbol\hat{}}{\boldsymbol j}+u_3\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $\boldsymbol v=v_1\overset{\boldsymbol\hat{}}{\boldsymbol i}+v_2\overset{\boldsymbol\hat{}}{\boldsymbol j}+v_3\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dapat dihitung menggunakan rumus berikut:

$cos θ = = \frac{u \cdot v}{∣ u ∣ ∣ v ∣} \frac{u _{1} v _{1} + u _{2} v _{2} + u _{3} v _{3}}{( u _{1}^{2} + u _{2}^{2} + u _{3}^{2} ) ( v _{1}^{2} + v _{2}^{2} + v _{3}^{2} )}$

Sudut antara ${\boldsymbol v}_{\mathbf1}=\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}+\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $\frac12\left(\overset{\boldsymbol\hat{}}{\boldsymbol i}-\overset{\mathit\hat{}}{\boldsymbol j}-\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)$ , yaitu:

$\begin{array}{rcl}\cos\;\theta_1&=&\frac{\left({\displaystyle\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}+\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}}\right)\cdot\left({\displaystyle\frac12\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac12\overset{\boldsymbol\hat{}}{\boldsymbol j}-\frac12\overset{\boldsymbol\hat{}}{\boldsymbol k}}\right)}{\left|{\displaystyle\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}-\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}+\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}}\right|\left|\mathbf{\displaystyle\frac{\mathrm1}{\mathrm2}\overset{\boldsymbol{\mathit\hat{}}}{\boldsymbol i}\mathrm-\frac{\mathrm1}{\mathrm2}\overset{\boldsymbol{\mathit\hat{}}}{\boldsymbol j}\mathrm-\frac{\mathrm1}{\mathrm2}\overset{\boldsymbol{\mathit\hat{}}}{\boldsymbol k}}\right|}\\&=&\frac{\left({\displaystyle\frac4{13}}\right)\left({\displaystyle\frac12}\right)+\left(-{\displaystyle\frac{12}{13}}\right)\left(-{\displaystyle\frac12}\right)+\left({\displaystyle\frac3{13}}\right)\left(-{\displaystyle\frac12}\right)}{\sqrt{\left(1\right)\left(\left({\displaystyle\frac12}\right)^2+\left({\displaystyle\frac12}\right)^2+\left({\displaystyle\frac12}\right)^2\right)}}\\&=&\frac{{\displaystyle\frac4{26}}+{\displaystyle\frac{12}{26}}-{\displaystyle\frac3{26}}}{\sqrt{3\times\left({\displaystyle\frac12}\right)^2}}\\&=&\frac{\displaystyle\frac{13}{26}}{{\displaystyle\frac12}\sqrt2}\\&=&\frac{\cancel{\displaystyle\frac12}}{\cancel{\displaystyle\frac12}\sqrt3}\\&=&\frac1{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\\&=&\frac13\sqrt3\\\theta_1&=&\cos^{-1}\;\left(\frac13\sqrt3\right)\\&=&54,74^\circ\end{array}$

Sudut antara ${\boldsymbol v}_{\mathbf2}=-\frac4{13}\overset{\boldsymbol\hat{}}{\boldsymbol i}+\frac{12}{13}\overset{\boldsymbol\hat{}}{\boldsymbol j}-\frac3{13}\overset{\boldsymbol\hat{}}{\boldsymbol k}$ dan $\frac12\left(\overset{\boldsymbol\hat{}}{\boldsymbol i}-\overset{\mathit\hat{}}{\boldsymbol j}-\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)$ , yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta subscript 2 end cell equals cell fraction numerator open parentheses negative begin display style 4 over 13 end style bold italic i with bold hat on top plus begin display style 12 over 13 end style bold italic j with bold hat on top minus begin display style 3 over 13 end style bold italic k with bold hat on top close parentheses times open parentheses begin display style 1 half bold italic i with bold hat on top minus 1 half bold italic j with bold hat on top minus 1 half bold italic k with bold hat on top end style close parentheses over denominator open vertical bar begin display style negative 4 over 13 bold italic i with bold hat on top plus 12 over 13 bold italic j with bold hat on top minus 3 over 13 bold italic k with bold hat on top end style close vertical bar open vertical bar begin bold style begin display style straight 1 over straight 2 end style bold italic i with bold italic hat on top straight minus begin display style straight 1 over straight 2 end style bold italic j with bold italic hat on top straight minus begin display style straight 1 over straight 2 end style bold italic k with bold italic hat on top end style close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses begin display style negative 4 over 13 end style close parentheses open parentheses begin display style 1 half end style close parentheses plus open parentheses begin display style 12 over 13 end style close parentheses open parentheses negative begin display style 1 half end style close parentheses plus open parentheses begin display style negative 3 over 13 end style close parentheses open parentheses negative begin display style 1 half end style close parentheses over denominator square root of open parentheses 1 close parentheses open parentheses open parentheses begin display style 1 half end style close parentheses squared plus open parentheses begin display style 1 half end style close parentheses squared plus open parentheses begin display style 1 half end style close parentheses squared close parentheses end root end fraction end cell row blank equals cell fraction numerator negative begin display style 4 over 26 end style minus begin display style 12 over 26 end style plus begin display style 3 over 26 end style over denominator square root of 3 cross times open parentheses begin display style 1 half end style close parentheses squared end root end fraction end cell row blank equals cell fraction numerator begin display style negative 13 over 26 end style over denominator begin display style 1 half end style square root of 2 end fraction end cell row blank equals cell fraction numerator negative up diagonal strike begin display style 1 half end style end strike over denominator up diagonal strike begin display style 1 half end style end strike square root of 3 end fraction end cell row blank equals cell negative fraction numerator 1 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell negative 1 third square root of 3 end cell row cell theta subscript 2 end cell equals cell cos to the power of negative 1 end exponent space open parentheses negative 1 third square root of 3 close parentheses end cell row blank equals cell 125 comma 26 degree end cell end table$

Dengan demikian, sudut antara masing-masing vektor satuan ini dengan vektor $\frac12\left(\overset{\boldsymbol\hat{}}{\boldsymbol i}-\overset{\mathit\hat{}}{\boldsymbol j}-\overset{\boldsymbol\hat{}}{\boldsymbol k}\right)$ adalah $θ_{1} = 54, 7 4^{\circ}$ dan $θ_{2} = 125, 2 6^{\circ}$ .

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

3.5 (2 rating)

Herlita Lita

Pembahasan tidak lengkap

Pertanyaan serupa

Misal a = 3 i + j − 4 k dan b = − 3 i + j + 4 k . Jika v tegak lurus a dan tegak lurus b dan ∣ v ∣ = 1 , maka = ...

4.8

Jawaban terverifikasi

Diketahui vektor a = t i − 2 t j + 4 k dan b = t i − 4 j + 3 k . Jika a tegak lurus b dan ∣ ∣ b ∣ ∣ = 29 maka ∣ a ∣ = ...

5.0

Jawaban terverifikasi

Diketahui a = 2 i − 4 j + 3 k ; b = x i + z j + 4 k ; c = 5 i − 3 j + 2 k dan d = 2 i + z j + x k . Jika a tegak lurus b dan c tegak lurus d , maka ∣ ∣ a + b ∣ ∣ = ...

4.4

Jawaban terverifikasi

Tentukan suatu vektor yang besarnya 1 dan tegak lurus terhadap kedua vektor a = ( 2 , − 2 , 1 ) dan b = ( 2 , 3 , − 4 )

5.0

Jawaban terverifikasi

Tentukan suatu vektor yang besarnya 1 dan tegak lurus terhadap kedua vektor a = ( 2 , − 2 , 1 ) dan b = ( 2 , 3 , − 4 )

5.0

Jawaban terverifikasi