Soal-Soal Matematika/Vektor

Vektor

Posisi vektor

${\vec {a}}=(a_{1},a_{2})={\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}=a_{1}{\hat {i}}+a_{2}{\hat {j}}$

${\vec {a}}=(a_{1},a_{2},a_{3})={\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix}}=a_{1}{\hat {i}}+a_{2}{\hat {j}}+a_{3}{\hat {k}}$

Panjang vektor

Berada di $R^{2}$

Panjang vektor a dalam posisi $(a_{1},a_{2})$ adalah $\left|{\vec {a}}\right|={\sqrt {a_{1}^{2}+a_{2}^{2}}}$

Panjang vektor b dalam posisi $(b_{1},b_{2})$ adalah $\left|{\vec {b}}\right|={\sqrt {b_{1}^{2}+b_{2}^{2}}}$

Panjang vektor c dalam posisi $(a_{1},a_{2})$ dan $(b_{1},b_{2})$ adalah $\left|{\vec {c}}\right|={\sqrt {(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}}}$

Berada di $R^{3}$

Panjang vektor a dalam posisi $(a_{1},a_{2},a_{3})$ adalah $\left|{\vec {a}}\right|={\sqrt {a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}}$

Panjang vektor b dalam posisi $(b_{1},b_{2},b_{3})$ adalah $\left|{\vec {b}}\right|={\sqrt {b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}$

Panjang vektor c dalam posisi $(a_{1},a_{2},a_{3})$ dan $(b_{1},b_{2},b_{3})$ adalah $\left|{\vec {c}}\right|={\sqrt {(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}}}$

Jumlah dan selisih kedua vektor

$\left|{\vec {a}}\pm {\vec {b}}\right|={\sqrt {|{\vec {a}}|^{2}+|{\vec {b}}|^{2}\pm 2\cdot |{\vec {a}}|\cdot |{\vec {b}}|\cdot cosC}}$

Vektor satuan

${\hat {a}}={\frac {\vec {a}}{\left|{\vec {a}}\right|}}$

Operasi aljabar pada vektor

• Penjumlahan dan pengurangan

terdiri dari 2 aturan jenis yaitu aturan segitiga dan jajar genjang

${\vec {c}}={\vec {a}}+{\vec {b}}={\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+{\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}={\begin{pmatrix}{a_{1}+b_{1}}\\{a_{2}+b_{2}}\end{pmatrix}}$

${\vec {c}}={\vec {a}}-{\vec {b}}={\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}-{\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}={\begin{pmatrix}{a_{1}-b_{1}}\\{a_{2}-b_{2}}\end{pmatrix}}$

• Perkalian

1. skalar dengan vektor

Jika k skalar tak nol dan vektor ${\vec {a}}=a_{1}{\hat {i}}+a_{2}{\hat {j}}+a_{3}{\hat {k}}$ maka vektor $k{\vec {a}}=(ka_{1},ka_{2},ka_{3})$

1. titik dua vektor

Jika vektor ${\vec {a}}=a_{1}{\hat {i}}+a_{2}{\hat {j}}+a_{3}{\hat {k}}$ dan vektor ${\vec {b}}=b_{1}{\hat {i}}+b_{2}{\hat {j}}+b_{3}{\hat {k}}$ maka ${\vec {a}}\cdot {\vec {b}}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$

1. titik dua vektor dengan membentuk sudut

Jika ${\vec {a}}$ dan ${\vec {b}}$ vektor tak nol dan sudut $\alpha$ diantara vektor ${\vec {a}}$ dan ${\vec {b}}$ maka perkalian skalar vektor ${\vec {a}}$ dan ${\vec {b}}$ adalah ${\vec {a}}\cdot {\vec {b}}$ = $\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|cos\alpha$

1. silang dua vektor

Jika vektor ${\vec {a}}=a_{1}{\hat {i}}+a_{2}{\hat {j}}+a_{3}{\hat {k}}$ dan vektor ${\vec {b}}=b_{1}{\hat {i}}+b_{2}{\hat {j}}+b_{3}{\hat {k}}$ maka ${\vec {a}}\times {\vec {b}}=(a_{2}b_{3}{\hat {i}}+a_{3}b_{1}{\hat {j}}+a_{1}b_{2}{\hat {k}})-(a_{2}b_{1}{\hat {k}}+a_{3}b_{2}{\hat {i}}+a_{1}b_{3}{\hat {j}})$

$\left[{\begin{array}{rrr|rr}{\hat {i}}&{\hat {j}}&{\hat {k}}&{\hat {i}}&{\hat {j}}\\a_{1}&a_{2}&a_{3}&a_{1}&a_{2}\\b_{1}&b_{2}&b_{3}&b_{1}&b_{2}\\-&-&-+&+&+\\\end{array}}\right]$

1. silang dua vektor dengan membentuk sudut

Jika ${\vec {a}}$ dan ${\vec {b}}$ vektor tak nol dan sudut $\alpha$ diantara vektor ${\vec {a}}$ dan ${\vec {b}}$ maka perkalian skalar vektor ${\vec {a}}$ dan ${\vec {b}}$ adalah ${\vec {a}}\times {\vec {b}}$ = $\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|sin\alpha$

Sifat operasi aljabar pada vektor

1. ${\vec {a}}+{\vec {b}}={\vec {b}}+{\vec {a}}$
2. $({\vec {a}}+{\vec {b}})+{\vec {c}}={\vec {a}}+({\vec {b}}+{\vec {c}})$
3. ${\vec {a}}+0=0+{\vec {a}}$
4. $k({\vec {a}}+{\vec {b}})=k{\vec {a}}+k{\vec {b}}$
5. $(k+l){\vec {a}}=k{\vec {a}}+l{\vec {a}}$
6. ${\vec {a}}+(-{\vec {a}})=0$
7. ${\vec {a}}\cdot {\vec {b}}={\vec {b}}\cdot {\vec {a}}$
8. $({\vec {a}}\cdot {\vec {b}})\cdot {\vec {c}}={\vec {a}}\cdot ({\vec {b}}\cdot {\vec {c}})$
9. ${\vec {a}}\cdot 1=1\cdot {\vec {a}}$
10. $k({\vec {a}}\cdot {\vec {b}})=k{\vec {a}}\cdot {\vec {b}}={\vec {a}}\cdot k{\vec {b}}$
11. $(k\cdot l){\vec {a}}=k(l\cdot {\vec {a}})$
12. ${\vec {a}}\cdot {\vec {a}}=\left|{\vec {a}}\right|^{2}$
13. ${\vec {a}}\times {\vec {b}}\neq {\vec {b}}\times {\vec {a}}$
14. ${\vec {a}}\times {\vec {b}}=-({\vec {b}}\times {\vec {a}})$
15. $({\vec {a}}\times {\vec {b}})\times {\vec {c}}\neq {\vec {a}}\times ({\vec {b}}\times {\vec {c}})$
16. $k({\vec {a}}\times {\vec {b}})=k{\vec {a}}\times {\vec {b}}={\vec {a}}\times k{\vec {b}}$
17. ${\vec {a}}\cdot ({\vec {b}}\times {\vec {c}})={\vec {b}}\cdot ({\vec {c}}\times {\vec {a}})={\vec {c}}\cdot ({\vec {a}}\times {\vec {b}})$
18. ${\vec {a}}\times ({\vec {b}}+{\vec {c}})={\vec {a}}\times {\vec {b}}+{\vec {a}}\times {\vec {c}}$

Hubungan vektor dengan vektor lain

• Perkalian titik

Saling tegak lurus

Jika tegak lurus antara vektor ${\vec {a}}$ dengan vektor ${\vec {b}}$ maka

${\vec {a}}\cdot {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|\cos {90}^{\circ }$

${\vec {a}}\cdot {\vec {b}}=0$

Sejajar

Jika vektor ${\vec {a}}$ sejajar dengan vektor ${\vec {b}}$ maka

${\vec {a}}\cdot {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|\cos {0}^{\circ }$

${\vec {a}}\cdot {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|$

${\vec {a}}\cdot {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|\cos {180}^{\circ }$

${\vec {a}}\cdot {\vec {b}}=-\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|$

• Perkalian silang

Saling tegak lurus

Jika tegak lurus antara vektor ${\vec {a}}$ dengan vektor ${\vec {b}}$ maka

${\vec {a}}\times {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|\sin {90}^{\circ }$

${\vec {a}}\times {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|$

${\vec {a}}\times {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|\sin {270}^{\circ }$

${\vec {a}}\times {\vec {b}}=-\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|$

Jika $\beta >{90}^{\circ }$ maka dua vektor tersebut searah

Jika $\beta <{90}^{\circ }$ maka vektor saling berlawanan arah

Sejajar

Jika vektor ${\vec {a}}$ sejajar dengan vektor ${\vec {b}}$ maka

${\vec {a}}\times {\vec {b}}=\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|\sin {0}^{\circ }$

${\vec {a}}\times {\vec {b}}=0$

Sudut dua vektor

Jika vektor ${\vec {a}}$ dan vektor ${\vec {b}}$ sudut yang dapat dibentuk dari kedua vektor tersebut adalah $cos\alpha ={\frac {{\vec {a}}\cdot {\vec {b}}}{\left|{\vec {a}}\right|\cdot \left|{\vec {b}}\right|}}$

Panjang proyeksi dan proyeksi vektor

Panjang proyeksi vektor (proyeksi skalar ortogonal) ${\vec {a}}$ pada vektor ${\vec {b}}$ adalah $\left|{\vec {c}}\right|={\frac {{\vec {a}}\cdot {\vec {b}}}{\left|{\vec {b}}\right|}}$

Proyeksi vektor (proyeksi vektor ortogonal) ${\vec {a}}$ pada vektor ${\vec {b}}$ adalah ${\vec {c}}={\frac {{\vec {a}}\cdot {\vec {b}}}{\left|{\vec {b}}\right|^{2}}}\cdot {\vec {b}}$

Metode

perhatikan garis-garis sesuai dengan arah panahnya dalam bentuk rumus seperti ${\vec {AB}},{\vec {BA}},dll$ jika tidak ada tulisan O maka dianggap AB = b-a kecuali keterangan tertulis.

segitiga

${\vec {R}}={\vec {a}}+{\vec {b}}$

$\left|{\vec {AB}}\right|=\left|{\vec {OB}}\right|-\left|{\vec {OA}}\right|={\vec {b}}-{\vec {a}}$

$\left|{\vec {AB}}\right|=-\left|{\vec {BA}}\right|$

jajar genjang

${\vec {R}}={\vec {a}}+{\vec {b}}={\vec {b}}+{\vec {a}}$

$\left|{\vec {BA}}\right|=\left|{\vec {OA}}\right|-\left|{\vec {OB}}\right|={\vec {a}}-{\vec {b}}$

${\vec {R}}=|{\vec {a}}\pm {\vec {b}}|={\sqrt {|{\vec {a}}|^{2}+|{\vec {b}}|^{2}\pm 2\cdot {\vec {a}}\cdot {\vec {b}}\cdot cosC}}$

Titik segaris (kolinear)

Titik A, B dan C dikatakan segaris memenuhi persamaan sebagai berikut:

${\vec {AB}}=k{\vec {BC}},{\vec {BC}}=k{\vec {AC}},{\vec {AC}}=k{\vec {AB}}$

Perbandingan

Aturan jajar genjang

Posisi vektor

${\vec {N}}={\frac {ms+nr}{m+n}}$

Berada di $R^{2}$

${\vec {N}}=({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}})$

Berada di $R^{3}$

${\vec {N}}=({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}},{\frac {mz_{2}+nz_{1}}{m+n}})$

Satu garis

• Perbandingan posisi dalam adalah m:n

Posisi vektor

${\vec {N}}={\frac {ms+nr}{m+n}}$

Berada di $R^{2}$

${\vec {N}}=({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}})$

Berada di $R^{3}$

${\vec {N}}=({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}},{\frac {mz_{2}+nz_{1}}{m+n}})$

• Perbandingan posisi luar adalah m:-n

Posisi vektor

${\vec {N}}={\frac {ms-nr}{m-n}}$

Berada di $R^{2}$

${\vec {N}}=({\frac {mx_{2}-nx_{1}}{m-n}},{\frac {my_{2}-ny_{1}}{m-n}})$

Berada di $R^{3}$

${\vec {N}}=({\frac {mx_{2}-nx_{1}}{m-n}},{\frac {my_{2}-ny_{1}}{m-n}},{\frac {mz_{2}-nz_{1}}{m-n}})$

Luas

1. Luas segitiga

• misalkan vektor A (x1,y1,z1) dan B (x2,y2,z2) dengan sudut lancip

${\begin{aligned}L={\frac {1}{2}}\cdot {\begin{array}{rrr|r}x_{1}&y_{1}&z_{1}&x_{1}\\x_{2}&y_{2}&z_{2}&x_{2}\\-&-+&-+&+\\\end{array}}\\={\frac {1}{2}}\cdot |(x_{1}y_{2}{\hat {k}}+y_{1}z_{2}{\hat {i}}+z_{1}x_{2}{\hat {j}}-(x_{2}y_{1}{\hat {k}}+y_{2}z_{1}{\hat {i}}+z_{2}x_{1}{\hat {j}}))|\\={\frac {1}{2}}\cdot |((y_{1}z_{2}-y_{2}z_{1}){\hat {i}}+(z_{1}x_{2}-z_{2}x_{1}){\hat {j}}+(x_{1}y_{2}-x_{2}y_{1}){\hat {k}})|\\={\frac {1}{2}}{\sqrt {(y_{1}z_{2}-y_{2}z_{1})^{2}+(z_{1}x_{2}-z_{2}x_{1})^{2}+(x_{1}y_{2}-x_{2}y_{1})^{2}}}\\\end{aligned}}$

• misalkan vektor A (x1,y1,z1), B (x2,y2,z2) dan C(x3,y3,z3) dengan sudut lancip

misalkan AB dan AC

$AB=B-A=(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})=(x_{4},y_{4},z_{4})$

$AC=C-A=(x_{3}-x_{1},y_{3}-y_{1},z_{3}-z_{1})=(x_{5},y_{5},z_{5})$

${\begin{aligned}L={\frac {1}{2}}\cdot {\begin{array}{rrr|r}x_{4}&y_{4}&z_{4}&x_{4}\\x_{5}&y_{5}&z_{5}&x_{5}\\-&-+&-+&+\\\end{array}}\\={\frac {1}{2}}\cdot |(x_{4}y_{5}{\hat {k}}+y_{4}z_{5}{\hat {i}}+z_{4}x_{5}{\hat {j}}-(x_{5}y_{4}{\hat {k}}+y_{5}z_{4}{\hat {i}}+z_{5}x_{4}{\hat {j}}))|\\={\frac {1}{2}}\cdot |((y_{4}z_{5}-y_{5}z_{4}){\hat {i}}+(z_{4}x_{5}-z_{5}x_{4}){\hat {j}}+(x_{4}y_{5}-x_{5}y_{4}){\hat {k}})|\\={\frac {1}{2}}{\sqrt {(y_{4}z_{5}-y_{5}z_{4})^{2}+(z_{4}x_{5}-z_{5}x_{4})^{2}+(x_{4}y_{5}-x_{5}y_{4})^{2}}}\\\end{aligned}}$

1. Luas jajar genjang

misalkan vektor A (x1,y1,z1) dan B (x2,y2,z2) dengan sudut lancip

${\begin{aligned}L={\begin{array}{rrr|r}x_{1}&y_{1}&z_{1}&x_{1}\\x_{2}&y_{2}&z_{2}&x_{2}\\-&-+&-+&+\\\end{array}}\\=|(x_{1}y_{2}{\hat {k}}+y_{1}z_{2}{\hat {i}}+z_{1}x_{2}{\hat {j}}-(x_{2}y_{1}{\hat {k}}+y_{2}z_{1}{\hat {i}}+z_{2}x_{1}{\hat {j}}))|\\=|((y_{1}z_{2}-z_{1}y_{2}){\hat {i}}+(z_{1}x_{2}-y_{2}z_{1}){\hat {j}}+(x_{1}y_{2}-x_{2}y_{1}){\hat {k}})|\\={\sqrt {(y_{1}z_{2}-z_{1}y_{2})^{2}+(z_{1}x_{2}-y_{2}z_{1})^{2}+(x_{1}y_{2}-x_{2}y_{1})^{2}}}\\\end{aligned}}$

• misalkan vektor A (x1,y1,z1), B (x2,y2,z2) dan C(x3,y3,z3) dengan sudut lancip

misalkan AB dan AC

$AB=B-A=(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})=(x_{4},y_{4},z_{4})$

$AC=C-A=(x_{3}-x_{1},y_{3}-y_{1},z_{3}-z_{1})=(x_{5},y_{5},z_{5})$

${\begin{aligned}L={\frac {1}{2}}\cdot {\begin{array}{rrr|r}x_{4}&y_{4}&z_{4}&x_{4}\\x_{5}&y_{5}&z_{5}&x_{5}\\-&-+&-+&+\\\end{array}}\\=|(x_{4}y_{5}{\hat {k}}+y_{4}z_{5}{\hat {i}}+z_{4}x_{5}{\hat {j}}-(x_{5}y_{4}{\hat {k}}+y_{5}z_{4}{\hat {i}}+z_{5}x_{4}{\hat {j}}))|\\=|((y_{4}z_{5}-y_{5}z_{4}){\hat {i}}+(z_{4}x_{5}-z_{5}x_{4}){\hat {j}}+(x_{4}y_{5}-x_{5}y_{4}){\hat {k}})|\\={\sqrt {(y_{4}z_{5}-y_{5}z_{4})^{2}+(z_{4}x_{5}-z_{5}x_{4})^{2}+(x_{4}y_{5}-x_{5}y_{4})^{2}}}\\\end{aligned}}$

contoh

1. Titik ${\vec {a}}=-3{\vec {i}}-2{\vec {j}}+4{\vec {k}}$ dan ${\vec {b}}=6{\vec {i}}+6{\vec {j}}+{\vec {k}}$. tentukan:

• ${\vec {a}}+{\vec {b}}$
• ${\vec {a}}-{\vec {b}}$
• ${\vec {a}}\cdot {\vec {b}}$
• ${\vec {a}}\times {\vec {b}}$
• panjang vektor ${\vec {a}}{\text{ dan }}{\vec {b}}$
• vektor satuan pada vektor ${\hat {b}}$
• panjang proyeksi vektor ${\vec {a}}$ pada vektor ${\vec {b}}$
• proyeksi vektor ${\vec {a}}$ pada vektor ${\vec {b}}$

Jawaban

$${\begin{aligned}*{\vec {c}}&={\vec {a}}+{\vec {b}}\\&=(-3,-2,4)+(6,6,1)\\&=(-3+6,-2+6,4+1)\\&=(3,4,5)\\*{\vec {c}}&={\vec {a}}-{\vec {b}}\\&=(-3,-2,4)-(6,6,1)\\&=(-3-6,-2-6,4-1)\\&=(-9,-8,3)\\*{\vec {a}}\cdot {\vec {b}}\\&=(-3,-2,4)\cdot (6,6,1)\\&=(-3\cdot 6)+(-2\cdot 6)+(4\cdot 1)\\&=-18-12+4\\&=-26\\*{\vec {a}}\times {\vec {b}}\\{\begin{array}{rrr|rr}i&j&k&i&j\\-3&-2&4&-3&-2\\6&6&1&6&6\\-&-&-+&+&+\\\end{array}}\\&=(-3\cdot 6)k+(-2\cdot 1)i+(4\cdot 6)j-(-2\cdot 6)k-(4\cdot 6)i-(-3\cdot 1)j\\&=-18k-2i+24j+12k-24i+3j\\&=-26i+27j-6k\\*|{\vec {c}}|&={\sqrt {(6-(-3))^{2}+(6-(-2))^{2}+(1-4)^{2}}}\\&={\sqrt {9^{2}+8^{2}+(-3)^{2}}}\\&={\sqrt {81+64+9}}\\&={\sqrt {145}}\\*{\hat {b}}&={\frac {\vec {b}}{|{\vec {b}}|}}\\&={\frac {(6,6,1)}{\sqrt {6^{2}+6^{2}+1^{2}}}}\\&={\frac {(6,6,1)}{\sqrt {36+36+1}}}\\&={\frac {(6,6,1)}{\sqrt {73}}}\\*|{\vec {c}}|&={\frac {{\vec {a}}\cdot {\vec {b}}}{|{\vec {b}}|}}\\&={\frac {-26}{\sqrt {73}}}\\*{\vec {c}}&={\frac {{\vec {a}}\cdot {\vec {b}}}{|{\vec {b}}|^{2}}}\cdot {\vec {b}}\\&={\frac {-26}{({\sqrt {73}})^{2}}}(6,6,1)\\&=-{\frac {26}{73}}(6,6,1)\\\end{aligned}}$$

1. Titik ${\vec {a}}=-3{\vec {i}}+2{\vec {j}}-7{\vec {k}}$ dan ${\vec {b}}=x{\vec {i}}+5{\vec {j}}+{\vec {k}}$. tentukan:

• nilai a jika vektor ${\vec {a}}$ tegak lurus terhadap vektor ${\vec {b}}$
• nilai ${\vec {a}}\times {\vec {b}}$

Jawaban

$${\begin{aligned}*{\vec {a}}\cdot {\vec {b}}&=0\\(-3,2,-7)\cdot (x,5,1)&=0\\(-3\cdot x)+(2\cdot 5)+(-7\cdot 1)&=0\\-3x+10-7&=0\\-3x&=-3\\x&=1\\*{\vec {a}}\times {\vec {b}}\\{\begin{array}{rrr|rr}i&j&k&i&j\\-3&2&-7&-3&2\\1&5&1&1&5\\-&-&-+&+&+\\\end{array}}\\&=(-3\cdot 5)k+(2\cdot 1)i+(-7\cdot 1)j-(2\cdot 1)k-(-7\cdot 5)i-(-3\cdot 1)j\\&=-15k+2i-7j-2k+35i+3j\\&=37i+10j-17k\\\end{aligned}}$$

1. Titik ${\vec {a}}=-3{\vec {i}}+2{\vec {j}}-7{\vec {k}}$ dan ${\vec {b}}=x{\vec {i}}+5{\vec {j}}+{\vec {k}}$. tentukan jumlah nilai x jika vektor ${\vec {w}}$ tegak lurus terhadap vektor ${\vec {a}}$ dan ${\vec {b}}$ dan panjang vektor ${\vec {w}}$ adalah 2!

Jawaban

$${\begin{aligned}*{\text{ cara 1 }}\\{\vec {a}}\cdot {\vec {b}}&=|{\vec {a}}|\cdot |{\vec {b}}|cos\alpha \\(-3,2,-7)\cdot (x,5,1)&={\sqrt {(-3)^{2}+2^{2}+(-7)^{2}}}\cdot {\sqrt {x^{2}+5^{2}+1^{2}}}cos180^{\circ }\\(-3\cdot x)+(2\cdot 5)+(-7\cdot 1)&={\sqrt {62}}{\sqrt {x^{2}+26}}(-1)\\-3x+10-7&=-{\sqrt {62(x^{2}+26)}}\\-3x+3&=-{\sqrt {62(x^{2}+26)}}\\(-3x+3)^{2}&=62x^{2}+1.612\\9x^{2}-18x+9&=62x^{2}+1.612\\53x^{2}+18x+1.603&=0\\x_{1}+x_{2}&=-{\frac {18}{53}}\\*{\text{ cara 2 }}\\{\text{ misalkan }}{\vec {w}}=(a,b,c)\\{\vec {a}}\cdot {\vec {w}}&=0\\(-3,2,-7)\cdot (a,b,c)&=0\\(-3\cdot a)+(2\cdot b)+(-7\cdot c)&=0\\-3a+2b-7c&=0\,(1)\\{\vec {b}}\cdot {\vec {w}}&=0\\(x,5,1)\cdot (a,b,c)&=0\\(x\cdot a)+(5\cdot b)+(1\cdot c)&=0\\xa+5b+c&=0\,(2)\\a^{2}+b^{2}+c^{2}&=2^{2}\\a^{2}+b^{2}+c^{2}&=4\\{\text{ eliminasi persamaan 1 dan 2 }}\\(-3+7x)a+37b&=0\\b&={\frac {3a-7xa}{37}}\\c&=-xa-5b\\&=-xa-5({\frac {3a-7xa}{37}})\\&={\frac {-37xa-15a+35xa}{37}}\\a^{2}+b^{2}+c^{2}&=4\\a^{2}+({\frac {3a-7xa}{37}})^{2}+({\frac {-37xa-15a+35xa}{37}})^{2}&=4\\a^{2}+{\frac {(3a-7xa)^{2}}{37^{2}}}+{\frac {(-2xa-15a)^{2}}{37^{2}}}&=4(37^{2})\\37^{2}a^{2}+9a^{2}-42xa^{2}+49x^{2}a^{2}+(-2)^{2}x^{2}a^{2}+60xa^{2}-15^{2}a^{2}&=4(37^{2})\\1.369a^{2}+9a^{2}-42xa^{2}+49x^{2}a^{2}+4x^{2}a^{2}+60xa^{2}-225a^{2}&=5.476\\53a^{2}x^{2}+18a^{2}x+1.153a^{2}-5.476&=0\\x_{1}+x_{2}&=-{\frac {18a^{2}}{53a^{2}}}=-{\frac {18}{53}}\\\end{aligned}}$$

1. Diketahui segitiga ABC dengan titik A (1,5,1), B (3,4,1) dan C (2,2,1) maka berapa besar sudut ABC?

Jawaban

$${\begin{aligned}*{\text{ cara 1 }}\\AB&={\sqrt {(1-3)^{2}+(5-4)^{2}+(1-1)^{2}}}\\&={\sqrt {4+1+0}}\\&={\sqrt {5}}\\AC&={\sqrt {(1-2)^{2}+(5-2)^{2}+(1-1)^{2}}}\\&={\sqrt {1+9+0}}\\&={\sqrt {10}}\\BC&={\sqrt {(3-2)^{2}+(4-2)^{2}+(1-1)^{2}}}\\&={\sqrt {1+4+0}}\\&={\sqrt {5}}\\AC^{2}&=AB^{2}+BC^{2}-2\cdot AB\cdot AC\cdot cos\alpha \\({\sqrt {10}})^{2}&=({\sqrt {5}})^{2}+({\sqrt {5}})^{2}-2\cdot {\sqrt {5}}\cdot {\sqrt {5}}\cdot cos\alpha \\10&=5+5-10\cos \alpha \\10&=10-10\cos \alpha \\10\cos \alpha &=0\\cos\alpha &=0\\\alpha &=90^{\circ }\\*{\text{ cara 2 }}\\{\vec {BA}}&=(1-3,5-4,1-1)\\&=(-2,1,0)\\|{\vec {BA}}|&={\sqrt {(-2)^{2}+1^{2}+0^{2}}}\\&={\sqrt {4+1+0}}\\&={\sqrt {5}}\\{\vec {BC}}&=(2-3,2-4,1-1)\\&=(-1,-2,0)\\|{\vec {BC}}|&={\sqrt {(-1)^{2}+(-2)^{2}+0^{2}}}\\&={\sqrt {1+4+0}}\\&={\sqrt {5}}\\cos\alpha &={\frac {{\vec {BA}}\cdot {\vec {BC}}}{|{\vec {BA}}|\cdot |{\vec {BC}}|}}\\&={\frac {-2(-1)+1(-2)+0(0)}{{\sqrt {5}}\cdot {\sqrt {5}}}}\\&={\frac {2-2+0}{5}}\\&={\frac {0}{5}}\\&=0\\\alpha &=90^{\circ }\\\end{aligned}}$$

1. Berapa luas segitiga titik A (3,1,2), B (4,3,0) dan C (1,2,5)?

Jawaban

$${\begin{aligned}{\vec {AB}}&={\vec {B}}-{\vec {A}}=(4,3,0)-(3,1,2)=(1,2,-2)\\{\vec {BC}}&={\vec {C}}-{\vec {B}}=(1,2,5)-(4,3,0)=(-3,-1,5)\\{\vec {AC}}&={\vec {C}}-{\vec {A}}=(1,2,5)-(3,1,2)=(-2,1,3)\\*{\text{cara 1}}\\AB{\text{ dan }}BC\\L&={\frac {1}{2}}\cdot {\begin{array}{rrr|r}1&2&-2&1\\-3&-1&5&-3\\-&-+&-+&+\\\end{array}}\\&={\frac {1}{2}}\cdot |(1(-1){\hat {k}}+2(5){\hat {i}}+(-2)(-3){\hat {j}}-((-3)(2){\hat {k}}+(-1)(-2){\hat {i}}+(5)1{\hat {j}}))|\\&={\frac {1}{2}}\cdot |(-{\hat {k}}+10{\hat {i}}+6{\hat {j}}-(-6{\hat {k}}+2{\hat {i}}+5{\hat {j}}))|\\&={\frac {1}{2}}\cdot |(8{\hat {i}}+{\hat {j}}+5{\hat {k}})|\\&={\frac {1}{2}}\cdot {\sqrt {8^{2}+1^{2}+5^{2}}}\\&={\frac {1}{2}}\cdot {\sqrt {64+1+25}}\\&={\frac {1}{2}}\cdot {\sqrt {90}}\\&={\frac {3}{2}}{\sqrt {10}}\\*{\text{cara 2}}\\AB{\text{ dan }}AC\\L&={\frac {1}{2}}\cdot {\begin{array}{rrr|r}1&2&-2&1\\-2&1&3&-2\\-&-+&-+&+\\\end{array}}\\&={\frac {1}{2}}\cdot |(1(1){\hat {k}}+2(3){\hat {i}}+(-2)(-2){\hat {j}}-((-2)(2){\hat {k}}+1(-2){\hat {i}}+(3)1{\hat {j}}))|\\&={\frac {1}{2}}\cdot |({\hat {k}}+6{\hat {i}}+4{\hat {j}}-(-4{\hat {k}}-2{\hat {i}}+3{\hat {j}}))|\\&={\frac {1}{2}}\cdot |(8{\hat {i}}+{\hat {j}}+5{\hat {k}})|\\&={\frac {1}{2}}\cdot {\sqrt {8^{2}+1^{2}+5^{2}}}\\&={\frac {1}{2}}\cdot {\sqrt {64+1+25}}\\&={\frac {1}{2}}\cdot {\sqrt {90}}\\&={\frac {3}{2}}{\sqrt {10}}\\*{\text{cara 3}}\\BC{\text{ dan }}AC\\L&={\frac {1}{2}}\cdot {\begin{array}{rrr|r}-3&-1&5&-3\\-2&1&3&-2\\-&-+&-+&+\\\end{array}}\\&={\frac {1}{2}}\cdot |((-3)1{\hat {k}}+(-1)(3){\hat {i}}+(5)(-2){\hat {j}}-((-2)(-1){\hat {k}}+1(5){\hat {i}}+(3)(-3){\hat {j}}))|\\&={\frac {1}{2}}\cdot |(-3{\hat {k}}-3{\hat {i}}-10{\hat {j}}-(2{\hat {k}}+5{\hat {i}}-9{\hat {j}}))|\\&={\frac {1}{2}}\cdot |(-8{\hat {i}}-{\hat {j}}-5{\hat {k}})|\\&={\frac {1}{2}}\cdot {\sqrt {(-8)^{2}+(-1)^{2}+(-5)^{2}}}\\&={\frac {1}{2}}\cdot {\sqrt {64+1+25}}\\&={\frac {1}{2}}\cdot {\sqrt {90}}\\&={\frac {3}{2}}{\sqrt {10}}\\\end{aligned}}$$

1. Buktikan bahwa titik A (-1,3,6), B (1,1,5) dan C (3,-1,4) adalah titik segaris (kolinear)!

Jawaban

$${\begin{aligned}*{\vec {AB}}&=k{\vec {BC}}\\(b-a)&=k(c-b)\\((1,1,5)-(-1,3,6))&=k((3,-1,4)-(1,1,5))\\(2,-2,-1)&=k(2,-2,-1)\\k&=1\\*{\vec {BC}}&=k{\vec {AC}}\\(c-b)&=k(c-a)\\((3,-1,4)-(1,1,5))&=k((3,-1,4)-(-1,3,6))\\(2,-2,-1)&=k(4,-4,-2)\\k&={\frac {1}{2}}\\*{\vec {AC}}&=k{\vec {AB}}\\(c-b)&=k(c-a)\\((3,-1,4)-(-1,3,6))&=k((1,1,5)-(-1,3,6))\\(4,-4,-2)&=k(2,-2,-1)\\k&=2\\{\text{terbukti. ketiga titik tersebut merupakan kolinear. }}\\\end{aligned}}$$

1. Diketahui vektor OA = -i+3j+k dan OB = -i-2j+6k. Titik P membagi ${\vec {AB}}$ di dalam dengan perbandingan 2:3 maka tentukan panjang vektor:

• ${\vec {PA}}$
• ${\vec {PB}}$

Jawaban

$${\begin{aligned}{\text{karena titik P berada di dalam maka }}{\vec {AP}}:{\vec {PB}}=2:3\\{\vec {O}}P&=({\frac {3(-1)+2(-1)}{5}},{\frac {3(3)+2(-2)}{5}},{\frac {3(1)+2(6)}{5}})\\&=({\frac {-3-2}{5}},{\frac {9-4}{5}},{\frac {3+12}{5}})\\&=(-1,1,3)\\*{\vec {PA}}&={\vec {OA}}-{\vec {OP}}\\&=(-1,3,1)-(-1,1,3)\\&=(0,2,-2)\\{\text{jadi }}{\vec {PA}}{\text{ adalah }}2j+2k\\*{\vec {PB}}&={\vec {OB}}-{\vec {OP}}\\&=(-1,-2,6)-(-1,1,3)\\&=(0,-3,3)\\{\text{jadi }}{\vec {PA}}{\text{ adalah }}-3j+3k\\\end{aligned}}$$

1. Diketahui vektor OA = 3i-j+5k dan OB = i+j-k. Titik P membagi ${\vec {AB}}$ di luar dengan perbandingan 1:3 maka tentukan panjang vektor:

• ${\vec {PA}}$
• ${\vec {PB}}$

Jawaban

$${\begin{aligned}{\text{karena titik P berada di luar maka }}{\vec {PA}}:{\vec {PB}}=1:3{\text{ atau }}{\vec {AP}}:{\vec {PB}}=-1:3\\{\vec {O}}P&=({\frac {3(3)-(1)}{2}},{\frac {3(-1)-(1)}{2}},{\frac {3(5)-(-1)}{2}})\\&=({\frac {9-1}{2}},{\frac {-3-1}{2}},{\frac {15+1}{2}})\\&=(4,-2,8)\\*{\vec {PA}}&={\vec {OA}}-{\vec {OP}}\\&=(3,-1,5)-(4,-2,8)\\&=(-1,1,-3)\\{\text{jadi }}{\vec {PA}}{\text{ adalah }}-i+j-3k\\*{\vec {PB}}&={\vec {OB}}-{\vec {OP}}\\&=(3,1,-1)-(-2,-2,8)\\&=(5,3,-9)\\{\text{jadi }}{\vec {PA}}{\text{ adalah }}5i+3j-9k\\\end{aligned}}$$

1. Diketahui vektor OA = i+j+2k dan OB = i+2j+3k. Titik P terletak pada garis AB sehingga $|{\vec {AP}}|=|{\vec {OB}}|$ maka tentukan:

• OA . AP
• OA . OP

Jawaban

$${\begin{aligned}{\vec {OA}}=i+j+2k,{\vec {OB}}=i+2j+3k\\{\vec {AB}}&={\vec {OB}}-{\vec {OA}}\\&=(1,2,3)-(1,1,2)\\&=(0,1,1)\\***{\text{cara 1}}\\|{\vec {A}}B|&={\sqrt {0^{2}+1^{2}+1^{2}}}\\&={\sqrt {2}}\\|{\vec {O}}B|&={\sqrt {1^{2}+2^{2}+3^{2}}}\\&={\sqrt {14}}\\{\text{karena }}|{\vec {A}}P|=|{\vec {O}}B|{\text{ maka }}|{\vec {A}}P|={\sqrt {14}}\\{\vec {A}}P&={\frac {|{\vec {A}}P|}{|{\vec {A}}B|}}\cdot {\vec {A}}B\\&={\frac {\sqrt {14}}{\sqrt {2}}}\cdot (0,1,1)\\&={\frac {\sqrt {14}}{\sqrt {2}}}\cdot (0,1,1)\\&={\sqrt {7}}(0,1,1)\\&=(0,{\sqrt {7}},{\sqrt {7}})\\{\vec {O}}A\cdot {\vec {A}}P&=(1,1,2)\cdot (0,{\sqrt {7}},{\sqrt {7}})\\&=0+{\sqrt {7}}+2{\sqrt {7}}\\&=3{\sqrt {7}}\\**{\text{cara 2}}\\{\vec {A}}P&=k\cdot {\vec {A}}B\\&=k(0,1,1)\\&=(0,k,k)\\|{\vec {A}}P|&=|{\vec {O}}B|\\{\sqrt {0+k^{2}+k^{2}}}&={\sqrt {1^{2}+2^{2}+3^{2}}}\\{\sqrt {2k^{2}}}&={\sqrt {14}}\\k&={\sqrt {7}}\\{\vec {A}}P&=(0,{\sqrt {7}},{\sqrt {7}})\\{\vec {O}}A\cdot {\vec {A}}P&=(1,1,2)\cdot (0,{\sqrt {7}},{\sqrt {7}})\\&=0+{\sqrt {7}}+2{\sqrt {7}}\\&=3{\sqrt {7}}\\**{\text{cara 3}}\\|{\vec {A}}B|&={\sqrt {0^{2}+1^{2}+1^{2}}}\\&={\sqrt {2}}\\|{\vec {O}}B|&={\sqrt {1^{2}+2^{2}+3^{2}}}\\&={\sqrt {14}}\\{\vec {O}}A\cdot {\vec {A}}B&=|{\vec {O}}A|\cdot |{\vec {A}}B|cos\alpha \\cos\alpha &={\frac {{\vec {O}}A\cdot {\vec {A}}B}{|{\vec {O}}A|\cdot |{\vec {A}}B|}}\\{\vec {O}}A\cdot {\vec {A}}P&=|{\vec {O}}A|\cdot |{\vec {A}}P|cos\alpha \\&=|{\vec {O}}A|\cdot |{\vec {A}}P|({\frac {{\vec {O}}A\cdot {\vec {A}}B}{|{\vec {O}}A|\cdot |{\vec {A}}B|}})\\&=|{\vec {A}}P|({\frac {{\vec {O}}A\cdot {\vec {A}}B}{|{\vec {A}}B|}})\\{\text{karena }}|{\vec {A}}P|=|{\vec {O}}B|\\&=|{\vec {O}}B|({\frac {{\vec {O}}A\cdot {\vec {A}}B}{|{\vec {A}}B|}})\\&={\sqrt {14}}({\frac {(1,1,2)\cdot (0,1,1)}{\sqrt {2}}})\\&={\sqrt {7}}(0+1+2)\\&=3{\sqrt {7}}\\*{\vec {A}}P&=k\cdot {\vec {A}}B\\&=k(0,1,1)\\&=(0,k,k)\\|{\vec {A}}P|&=|{\vec {O}}B|\\{\sqrt {0+k^{2}+k^{2}}}&={\sqrt {1^{2}+2^{2}+3^{2}}}\\{\sqrt {2k^{2}}}&={\sqrt {14}}\\k&={\sqrt {7}}\\{\vec {A}}P&=(0,{\sqrt {7}},{\sqrt {7}})\\{\vec {O}}P&={\vec {O}}A+{\vec {A}}P\\&=(1,1,2)+(0,{\sqrt {7}},{\sqrt {7}})\\&=(1,1+{\sqrt {7}},2+{\sqrt {7}})\\{\vec {O}}A\cdot {\vec {O}}P&=(1,1,2)(1,1+{\sqrt {7}},2+{\sqrt {7}})\\&=1+1+{\sqrt {7}}+2(2+{\sqrt {7}})\\&=6+3{\sqrt {7}}\\\end{aligned}}$$