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Exercise Zone : Vektor

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Berikut ini adalah kumpulan soal mengenai Vektor. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

StandarSNBTHOTS


No.

Diketahui A = (a, −2, 3), B = (2, −1, 1), dan C = (1, 5, c). Agar vektor \overline{AB} tegak lurus pada \overline{BC}, maka nilai a − 2c sama dengan ....
Ganesha Operation
ALTERNATIF PENYELESAIAN
\begin{aligned} \overline{AB}&=\begin{pmatrix}2\\-1\\1\end{pmatrix}-\begin{pmatrix}a\\-2\\3\end{pmatrix}\\ &=\begin{pmatrix}2-a\\1\\-2\end{pmatrix} \end{aligned} \begin{aligned} \overline{BC}&=\begin{pmatrix}1\\5\\c\end{pmatrix}-\begin{pmatrix}2\\-1\\1\end{pmatrix}\\ &=\begin{pmatrix}-1\\6\\c-1\end{pmatrix} \end{aligned} \overline{AB} tegak lurus pada \overline{BC} maka \begin{aligned} \overline{AB}\cdot\overline{BC}&=0\\ \begin{pmatrix}2-a\\1\\-2\end{pmatrix}\cdot\begin{pmatrix}-1\\6\\c-1\end{pmatrix}&=0\\ (2-a)(-1)+(1)(6)+(-2)(c-1)&=0\\ -2+a+6-2c+2&=0\\ a-2c&=\boxed{\boxed{-6}} \end{aligned}
Jadi, nilai a − 2c sama dengan −6.

No.

P1 = (5, −2, 1) dan P2 = (2, 4, 2) maka vektor 5\overrightarrow{P_1P_2} adalah....
  1. (−15, 30, 5)
  2. (−13, 10, 5)
  3. (13, 2, 11)
  1. (−13, 2, 1)
  2. (−9, 6, 3)
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} 5\overrightarrow{P_1P_2}&=5\left(\begin{pmatrix}2\\4\\2\end{pmatrix}-\begin{pmatrix}5\\-2\\1\end{pmatrix}\right)\\ &=5\begin{pmatrix}-3\\6\\1\end{pmatrix}\\ &=\begin{pmatrix}-15\\30\\5\end{pmatrix}\\ &=\boxed{\boxed{(-15,30,5)}} \end{aligned}
Jadi, 5\overrightarrow{P_1P_2}=(-15,30,5).
JAWAB: A

No.

Diketahui vektor {\vec{a}=(4,6)}, {\vec{b}=(3,4)} dan {\vec{c}=(p,0)}. Jika {\left|\vec{c}-\vec{a}\right|=10} maka cosinus sudut antara vector \vec{b} dan \vec{c} yang mungkin adalah
  1. \dfrac25
  2. \dfrac12
  3. \dfrac35
  1. \dfrac23
  2. \dfrac34
SUMBER
ALTERNATIF PENYELESAIAN

CARA BIASA

\begin{aligned} \left|\vec{c}-\vec{a}\right|&=10\\ \sqrt{(p-4)^2+(0-6)^2}&=10\\ \sqrt{p^2-8p+16+36}&=10\\ \sqrt{p^2-8p+52}&=10\\ p^2-8p+52&=100\\ p^2-8p-48&=0\\ (p+4)(p-12)&=0 \end{aligned} p = −4 atau p = 12

Untuk p = −4

\vec{c}=(-4,0) \begin{aligned} \cos(\vec{b},\vec{c})&=\dfrac{\vec{b}\cdot\vec{c}}{\left|\vec{b}\right|\left|\vec{c}\right|}\\ &=\dfrac{(3)(-4)+(4)(0)}{\sqrt{3^2+4^2}\sqrt{(-4)^2+0^2}}\\ &=\dfrac{-12+0}{\sqrt{9+16}\sqrt{16+0}}\\ &=\dfrac{-12}{\sqrt{25}\sqrt{16}}\\ &=\dfrac{-12}{(5)(4)}\\ &=\boxed{\boxed{-\dfrac35}} \end{aligned}

Untuk p = 12

\vec{c}=(12,0) \begin{aligned} \cos(\vec{b},\vec{c})&=\dfrac{\vec{b}\cdot\vec{c}}{\left|\vec{b}\right|\left|\vec{c}\right|}\\ &=\dfrac{(3)(12)+(4)(0)}{\sqrt{3^2+4^2}\sqrt{12^2+0^2}}\\ &=\dfrac{36+0}{\sqrt{9+16}\sqrt{144+0}}\\ &=\dfrac{36}{\sqrt{25}\sqrt{144}}\\ &=\dfrac{36}{(5)(12)}\\ &=\boxed{\boxed{\dfrac35}} \end{aligned}

CARA CEPAT

\begin{aligned} \cos(\vec{b},\vec{c})&=\dfrac{\vec{b}\cdot\vec{c}}{\left|\vec{b}\right|\left|\vec{c}\right|}\\ &=\dfrac{(3)(p)+(4)(0)}{\sqrt{3^2+4^2}\sqrt{p^2+0^2}}\\ &=\dfrac{3p+0}{\sqrt{9+16}\sqrt{p^2+0}}\\ &=\dfrac{3p}{\sqrt{25}\sqrt{p^2}}\\ &=\dfrac{3p}{\sqrt{25}|p|}\\ &=\boxed{\boxed{\pm\dfrac35}} \end{aligned}
Jadi, cosinus sudut antara vector \vec{b} dan \vec{c} yang mungkin adalah \dfrac35.
JAWAB: C

No.

Diketahui \left|\vec{a}\right|=4, \left|\vec{b}\right|=5 serta \left|\vec{a}+\vec{b}\right|=6, tentukan nilai dari \left|\vec{a}-\vec{b}\right|
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \left|\vec{a}+\vec{b}\right|^2+\left|\vec{a}-\vec{b}\right|^2&=\left|\vec{a}\right|^2+\left|\vec{b}\right|^2\\ 6^2+\left|\vec{a}-\vec{b}\right|^2&=4^2+5^2\\ 36+\left|\vec{a}-\vec{b}\right|^2&=16+25\\ \left|\vec{a}-\vec{b}\right|^2&=5\\ \left|\vec{a}-\vec{b}\right|&=\boxed{\boxed{\sqrt5}} \end{aligned}
Jadi, \left|\vec{a}-\vec{b}\right|=\sqrt5.

No.

Diketahui \(\overrightarrow{PQ}=\pmatrix{1\\0}\) dan \(\overrightarrow{PR}=\pmatrix{2\\2}\). Jika {\overrightarrow{PS}=\dfrac14\overrightarrow{PQ}}, maka \overrightarrow{RS}=
Grup Telegram
ALTERNATIF PENYELESAIAN

CARA 1

\begin{aligned} \overrightarrow{PR}&=\pmatrix{2\\2}\\ \vec{r}-\vec{p}&=\pmatrix{2\\2}\\ \vec{p}-\vec{r}&=\pmatrix{-2\\-2} \end{aligned} \begin{aligned} \overrightarrow{PS}&=\dfrac14\overrightarrow{PQ}\\ \vec{s}-\vec{p}&=\dfrac14\pmatrix{1\\0}\\ &=\pmatrix{\dfrac14\\0} \end{aligned} \begin{aligned} \overrightarrow{RS}&=\vec{s}-\vec{r}\\ &=\vec{s}-\vec{p}+\vec{p}-\vec{r}\\ &=\pmatrix{\dfrac14\\0}+\pmatrix{-2\\-2}\\ &=\boxed{\boxed{\pmatrix{-\dfrac74\\-2}}} \end{aligned}

CARA 2

\begin{aligned} \overrightarrow{PS}&=\dfrac14\overrightarrow{PQ}\\ \overrightarrow{PR}+\overrightarrow{RS}&=\dfrac14\pmatrix{1\\0}\\ \pmatrix{2\\2}+\overrightarrow{RS}&=\pmatrix{\dfrac14\\0}\\ \overrightarrow{RS}&=\pmatrix{\dfrac14\\0}-\pmatrix{2\\2}\\ &=\boxed{\boxed{\pmatrix{-\dfrac74\\-2}}} \end{aligned}
Jadi, \(\overrightarrow{RS}=\pmatrix{-\dfrac74\\-2}\).

No.

Diberikan {\vec{a}=2\vec{i}+3\vec{j}-2\vec{k}} dan {\vec{b}=3\vec{i}-3\vec{j}-4\vec{k}}. Hasil \vec{a}\cdot\vec{b} adalah
  1. 3 satuan
  2. 4 satuan
  3. 5 satuan
  1. 8 satuan
  2. 10 satuan
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \vec{a}\cdot\vec{b}&=(2)(3)+(3)(-3)+(-2)(-4)\\ &=6-9+8\\ &=\boxed{\boxed{5}} \end{aligned}
Jadi, \vec{a}\cdot\vec{b}=5 satuan.
JAWAB: c

No.

Jika vektor a = 10i + 6j − 3k dan b = 8i + 3j + 3k serta c = a − b, maka vektor satuan yang searah dengan c adalah ....
  1. \dfrac67i+\dfrac27j+\dfrac37k
  2. \dfrac27i+\dfrac37j-\dfrac67k
  3. \dfrac67i-\dfrac37j+\dfrac67k
  1. \dfrac67i-\dfrac37j-\dfrac27k
  2. -\dfrac27i+\dfrac67j-\dfrac37k
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} c&=a-b\\ &=\left(10i+6j-3k\right)-\left(8i+3j+3k\right)\\ &=10i+6j-3k-8i-3j-3k\\ &=2i+3j-6k \end{aligned} \begin{aligned} |c|&=\sqrt{2^2+3^2+(-6)^2}\\ &=\sqrt{4+9+36}\\ &=\sqrt{49}\\ &=7 \end{aligned}
\begin{aligned} e_c&=\dfrac{c}{|c|}\\ &=\dfrac{2i+3j-6k}7\\ &=\boxed{\boxed{\dfrac27i+\dfrac37j-\dfrac67k}} \end{aligned}
Jadi, vektor satuan yang searah dengan c adalah \dfrac27i+\dfrac37j-\dfrac67k.
JAWAB: v

No.

Tentukan panjang vektor p = (3, 5, −4)
Brainly.co.id
ALTERNATIF PENYELESAIAN
\begin{aligned} |p|&=\sqrt{3^2+5^2+(-4)^2}\\ &=\sqrt{9+25+16}\\ &=\sqrt{50}\\ &=\boxed{\boxed{5\sqrt2}} \end{aligned}
Jadi, panjang vektor p = (3, 5, −4) adalah 5\sqrt2.

No.

Diketahui vektor {\vec{u}=x\vec{i}+3\vec{j}-4\vec{k}}, {\vec{v}=4\vec{i}+6\vec{j}+y\vec{k}} dan {\vec{w}=x\vec{i}-\vec{j}+(y+2)\vec{k}}. Jika {\vec{u}+\vec{v}=3\vec{w}}, tentukan nilai x dan y!
Grup Matematika Zone Id
ALTERNATIF PENYELESAIAN
\begin{aligned} \vec{u}+\vec{v}&=3\vec{w}\\ \pmatrix{x\\3\\-4}+\pmatrix{4\\6\\y}&=3\pmatrix{x\\3\\-4}\\ \pmatrix{x+4\\9\\y-4}&=\pmatrix{3x\\9\\-12} \end{aligned}
\begin{aligned} x+4&=3x\\ 4&=2x\\ x&=\boxed{\boxed{2}} \end{aligned} \begin{aligned} y-4&=-12\\ y&=\boxed{\boxed{-8}} \end{aligned}
Jadi, x = 2 dan y = −8.

No.

Diberikan tiga titik A, B, C dengan koordinat A(4, 5, 16); B(5, 2, 10); C(7, 10, 18). Pada ruas garis BC ditempatkan titik D sehingga BD : DC = 3 : 2. Koordinat titik D yang tepat adalah ....
  1. \(\left(\dfrac{11}5,\ \dfrac95,\ \dfrac{-6}5\right)\)
  2. \(\left(\dfrac{11}5,\ \dfrac95,\ \dfrac{6}5\right)\)
  3. \(\left(\dfrac65,\ \dfrac15,\ \dfrac{-14}5\right)\)
  1. \(\left(\dfrac{12}5,\ \dfrac{11}5,\ \dfrac{-6}5\right)\)
  2. \(\left(\dfrac{6}5,\ \dfrac{11}5,\ \dfrac{-9}5\right)\)
EDUVERSAL MATHEMATICS COMPETITION 2019
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \vec{d}&=\dfrac{2\begin{pmatrix}5\\2\\10\end{pmatrix}+3\begin{pmatrix}7\\10\\18\end{pmatrix}}{2+3}\\[4pt] &=\dfrac{\begin{pmatrix}10\\4\\20\end{pmatrix}+\begin{pmatrix}21\\30\\54\end{pmatrix}}5\\[4pt] &=\dfrac{\begin{pmatrix}31\\34\\74\end{pmatrix}}5\\[4pt] &=\begin{pmatrix}\dfrac{31}5\\\dfrac{34}5\\\dfrac{74}5\end{pmatrix} \end{aligned}\)
Jadi, koordinat titik D yang tepat adalah \(\left(\dfrac{31}5,\ \dfrac{34}5,\ \dfrac{74}5\right)\).
JAWAB: TIDAK ADA


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