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Diketahui p = 3i + 2j + k , q

Berikut ini adalah pertanyaan dari vivifnt pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Diketahui p = 3i + 2j + k , q = 2i – j + 2k , r = 3i + j – k, dan s = – 2i + 3j + 3k ,tentukan hasil perkalian vektor-vektor berikut: (Cross Product dan Dot Product)
a. q × (2p – r)
b. p . (q x r)
c. (p x 2q) . (r x s)
d. (p - s) × (q + r)
e. p x 2q x s
f. (p x 2q x r) . s

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jika diketahui : \vec{p}=a\vec{i}+b\vec{j}+c\vec{k}dan\vec{q}=x\vec{i}+y\vec{j}+z\vec{k}, maka :

Dot Product :

\purple{\huge{\vec{p}.\vec{q}=ax+by+cz}}

Cross Product :

\boxed{\purple{\begin{array}{ccc}\vec{p}\times \vec{q}=(bz-cy)\vec{i}\\+(cx-az)\vec{j}+(ay-bx)\vec{k}\end{array}}}

Cara menghitung cross product dua buah vektor sama dengan cara menghitung determinan matriks 3 × 3 seperti terlampir di gambar.

PERLU DIPERHATIKAN : \vec{p}.\vec{q}=\vec{q}.\vec{p}, tetapi \vec{p}\times \vec{q}\ne \vec{q}\times \vec{p}

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

\green{\huge{a.}}

\overrightarrow{2p}-\vec{r}=3\vec{i}+3\vec{j}+3\vec{k}[tex]\vec{q}\times \left(\overrightarrow{2p}-\vec{r}\right)=(-3-6)\vec{i}+(6-6)\vec{j}+(6-(-3))\vec{k}

\red{\huge{\begin{array}{ccc}\vec{q}\times \left(\overrightarrow{2p}-\vec{r}\right)=\\-9\vec{i}+9\vec{k}\end{array}}}

\\

\green{\huge{b.}}

\vec{q}\times \vec{r}=(1-2)\vec{i}+(6-(-2))\vec{j}+(2-(-3))\vec{k}=-\vec{i}+8\vec{j}+5\vec{k}

\vec{p}.\left(\vec{q}\times \vec{r}\right)=(3\times -1)+(2\times 8)+(1\times 5)=-3+16+5

\red{\huge{\vec{p}.\left(\vec{q}\times \vec{r}\right)=18}}

\\

\green{\huge{c.}}

\vec{p}\times \overrightarrow{2q}=(8-(-2))\vec{i}+(4-12)\vec{j}+(-6-8)\vec{k}=10\vec{i}-8\vec{j}-14\vec{k}

\vec{r}\times \vec{s}=(3-(-3))\vec{i}+(2-9)\vec{j}+(9-(-2))\vec{k}=6\vec{i}-7\vec{j}+11\vec{k}

\left(\vec{p}\times \overrightarrow{2q}\right).\left(\vec{r}\times \vec{s}\right)=(10\times 6)+(-8\times -7)+(-14\times 11)=60+56-154

\red{\huge{\begin{array}{ccc}\left(\vec{p}\times \overrightarrow{2q}\right).\left(\vec{r}\times \vec{s}\right)\\=-38\end{array}}}

\\

\green{\huge{d.}}

\vec{p}-\vec{s}=5\vec{i}-\vec{j}-2\vec{k}

\vec{q}+\vec{r}=5\vec{i}+\vec{k}

\left(\vec{p}-\vec{s}\right)\times \left(\vec{q}+\vec{r}\right)=(-1-0)\vec{i}+(-10-5)\vec{j}+(0-(-5))\vec{k}

\red{\huge{\begin{array}{ccc}\left(\vec{p}-\vec{s}\right)\times \left(\vec{q}+\vec{r}\right)\\=-\vec{i}-15\vec{j}+5\vec{k}\end{array}}}

\\

\green{\huge{e.}}

\vec{p}\times \overrightarrow{2q}=(8-(-2))\vec{i}+(4-12)\vec{j}+(-6-8)\vec{k}=10\vec{i}-8\vec{j}-14\vec{k}

\left(\vec{p}\times \overrightarrow{2q}\right)\times \vec{s}=(-24-(-42))\vec{i}+(28-30)\vec{j}+(30-16)\vec{k}

\red{\huge{\begin{array}{ccc}\vec{p}\times \overrightarrow{2q}\times \vec{s}\\=18\vec{i}-2\vec{j}+14\vec{k}\end{array}}}

\\

\green{\huge{f.}}

\vec{p}\times \overrightarrow{2q}\times \vec{r}=(8-(-14))\vec{i}+(-42-(-10))\vec{j}+(10-(-24))\vec{k}=22\vec{i}-32\vec{j}+34\vec{k}

\left(\vec{p}\times \overrightarrow{2q}\times \vec{r}\right).\vec{s}=(22\times -2)+(-32\times 3)+(34\times 3)=-44-96+102

\red{\huge{\begin{array}{ccc}\left(\vec{p}\times \overrightarrow{2q}\times \vec{r}\right).\vec{s}\\=-38\end{array}}}

Jika diketahui : [tex]\vec{p}=a\vec{i}+b\vec{j}+c\vec{k}[/tex] dan [tex]\vec{q}=x\vec{i}+y\vec{j}+z\vec{k}[/tex], maka :Dot Product :[tex]\purple{\huge{\vec{p}.\vec{q}=ax+by+cz}}[/tex]Cross Product :[tex]\boxed{\purple{\begin{array}{ccc}\vec{p}\times \vec{q}=(bz-cy)\vec{i}\\+(cx-az)\vec{j}+(ay-bx)\vec{k}\end{array}}}[/tex]Cara menghitung cross product dua buah vektor sama dengan cara menghitung determinan matriks 3 × 3 seperti terlampir di gambar.PERLU DIPERHATIKAN : [tex]\vec{p}.\vec{q}=\vec{q}.\vec{p}[/tex], tetapi [tex]\vec{p}\times \vec{q}\ne \vec{q}\times \vec{p}[/tex]•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••[tex]\green{\huge{a.}}[/tex][tex]\overrightarrow{2p}-\vec{r}=3\vec{i}+3\vec{j}+3\vec{k}[tex]\vec{q}\times \left(\overrightarrow{2p}-\vec{r}\right)=(-3-6)\vec{i}+(6-6)\vec{j}+(6-(-3))\vec{k}[/tex][tex]\red{\huge{\begin{array}{ccc}\vec{q}\times \left(\overrightarrow{2p}-\vec{r}\right)=\\-9\vec{i}+9\vec{k}\end{array}}}[/tex][tex]\\[/tex][tex]\green{\huge{b.}}[/tex][tex]\vec{q}\times \vec{r}=(1-2)\vec{i}+(6-(-2))\vec{j}+(2-(-3))\vec{k}[/tex][tex]=-\vec{i}+8\vec{j}+5\vec{k}[/tex][tex]\vec{p}.\left(\vec{q}\times \vec{r}\right)=(3\times -1)+(2\times 8)+(1\times 5)[/tex][tex]=-3+16+5[/tex][tex]\red{\huge{\vec{p}.\left(\vec{q}\times \vec{r}\right)=18}}[/tex][tex]\\[/tex][tex]\green{\huge{c.}}[/tex][tex]\vec{p}\times \overrightarrow{2q}=(8-(-2))\vec{i}+(4-12)\vec{j}+(-6-8)\vec{k}[/tex][tex]=10\vec{i}-8\vec{j}-14\vec{k}[/tex][tex]\vec{r}\times \vec{s}=(3-(-3))\vec{i}+(2-9)\vec{j}+(9-(-2))\vec{k}[/tex][tex]=6\vec{i}-7\vec{j}+11\vec{k}[/tex][tex]\left(\vec{p}\times \overrightarrow{2q}\right).\left(\vec{r}\times \vec{s}\right)=(10\times 6)+(-8\times -7)+(-14\times 11)[/tex][tex]=60+56-154[/tex][tex]\red{\huge{\begin{array}{ccc}\left(\vec{p}\times \overrightarrow{2q}\right).\left(\vec{r}\times \vec{s}\right)\\=-38\end{array}}}[/tex][tex]\\[/tex][tex]\green{\huge{d.}}[/tex][tex]\vec{p}-\vec{s}=5\vec{i}-\vec{j}-2\vec{k}[/tex][tex]\vec{q}+\vec{r}=5\vec{i}+\vec{k}[/tex][tex]\left(\vec{p}-\vec{s}\right)\times \left(\vec{q}+\vec{r}\right)=(-1-0)\vec{i}+(-10-5)\vec{j}+(0-(-5))\vec{k}[/tex][tex]\red{\huge{\begin{array}{ccc}\left(\vec{p}-\vec{s}\right)\times \left(\vec{q}+\vec{r}\right)\\=-\vec{i}-15\vec{j}+5\vec{k}\end{array}}}[/tex][tex]\\[/tex][tex]\green{\huge{e.}}[/tex][tex]\vec{p}\times \overrightarrow{2q}=(8-(-2))\vec{i}+(4-12)\vec{j}+(-6-8)\vec{k}[/tex][tex]=10\vec{i}-8\vec{j}-14\vec{k}[/tex][tex]\left(\vec{p}\times \overrightarrow{2q}\right)\times \vec{s}=(-24-(-42))\vec{i}+(28-30)\vec{j}+(30-16)\vec{k}[/tex][tex]\red{\huge{\begin{array}{ccc}\vec{p}\times \overrightarrow{2q}\times \vec{s}\\=18\vec{i}-2\vec{j}+14\vec{k}\end{array}}}[/tex][tex]\\[/tex][tex]\green{\huge{f.}}[/tex][tex]\vec{p}\times \overrightarrow{2q}\times \vec{r}=(8-(-14))\vec{i}+(-42-(-10))\vec{j}+(10-(-24))\vec{k}[/tex][tex]=22\vec{i}-32\vec{j}+34\vec{k}[/tex][tex]\left(\vec{p}\times \overrightarrow{2q}\times \vec{r}\right).\vec{s}=(22\times -2)+(-32\times 3)+(34\times 3)[/tex][tex]=-44-96+102[/tex][tex]\red{\huge{\begin{array}{ccc}\left(\vec{p}\times \overrightarrow{2q}\times \vec{r}\right).\vec{s}\\=-38\end{array}}}[/tex]

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Last Update: Sun, 11 Jul 21
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