tolong bantu nya kak ... terimakasih(urgent).... Question from @cahyarufika

July 2022 1 15 Report

tolong bantu nya kak ... terimakasih
(urgent)

$\begin{aligned}\sf4.\ &\vec{u}\cdot\vec{v}=\bf39\\\textsf{-----}&\textsf{---------------------------------}\\\sf5.\ &\textsf{Nilai tangen sudut antara}\\&\textsf{$\vec{u}$ dan $\vec{v}$ adalah }\bf\frac{1}{3}\\\textsf{-----}&\textsf{---------------------------------}\end{aligned}$

Pembahasan

Sudut Antara Dua Vektor

Nomor 4

Vektor $\vec{u}$ dan $\vec{v}$ di $R_2$ , dengan $\big|\vec{u}\big|=13$ , $\big|\vec{v}\big|=5$ , dan cos θ = 3/5.

Hasil perkalian titik (dot product) $\vec{u}\cdot\vec{v}$ adalah:

$\begin{aligned}\vec{u}\cdot\vec{v}&=\big|\vec{u}\big|\big|\vec{v}\big|\cos\theta\\&=13\cdot\cancel{5}\cdot\frac{3}{\cancel{5}}\\\therefore\ \vec{u}\cdot\vec{v}&=\boxed{\ \bf39\ }\end{aligned}$

$\blacksquare$

..................................

Nomor 5

Diketahui vektor-vektor:

$\vec{u}=\begin{pmatrix}2\\1\end{pmatrix}\,,\ \vec{v}=\begin{pmatrix}1\\1\end{pmatrix}$

Untuk menentukan nilai tangen sudut antara kedua vektor tersebut, kita dapat menggunakan beberapa cara.

Cara pertama: dengan perkalian silang dan perkalian titik.

Perkalian silang dua vektor di $R_2$ menghasilkan skalar, bukan vektor seperti pada perkalian silang dua vektor di $R_3$ .

$\begin{aligned}\tan\theta&=\frac{\sin\theta}{\cos\theta}=\frac{\left(\dfrac{\big|\vec{u}\times\vec{v}\big|}{\cancel{\big|\vec{u}\big|\cdot\big|\vec{v}\big|}}\right)}{\left(\dfrac{\vec{u}\cdot\vec{v}}{\cancel{\big|\vec{u}\big|\cdot\big|\vec{v}\big|}}\right)}\\\tan\theta&=\boxed{\ \frac{\big|\vec{u}\times\vec{v}\big|}{\vec{u}\cdot\vec{v}}\ }\\&=\frac{\det\begin{pmatrix}2&1\\1&1\end{pmatrix}}{\begin{pmatrix}2\\1\end{pmatrix}\cdot\begin{pmatrix}1\\1\end{pmatrix}}\\&=\frac{2-1}{2+1}\end{aligned}$

$\begin{aligned}\therefore\ \tan\theta&=\boxed{\ \bf\frac{1}{3}\ }\end{aligned}$

Cara kedua: dengan identitas trigonometri

$\begin{aligned}\tan\theta&=\pm\sqrt{\sec^2\theta-1}\\&=\pm\sqrt{\frac{1}{\cos^2\theta}-1}\\&=\pm\sqrt{\frac{1}{\left(\dfrac{\vec{u}\cdot\vec{v}}{\big|\vec{u}\big|\big|\vec{u}\big|}\right)^2}-1}\\&=\pm\sqrt{\frac{\left(\big|\vec{u}\big|\big|\vec{v}\big|\right)^2}{\left(\vec{u}\cdot\vec{v}\right)^2}-1}\\&=\pm\sqrt{\frac{\left(\sqrt{2^2+1^2}\sqrt{1^2+1^2}\right)^2}{\left(\begin{pmatrix}2\\1\end{pmatrix}\cdot\begin{pmatrix}1\\1\end{pmatrix}\right)^2}-1}\end{aligned}$

$\begin{aligned}\tan\theta&=\pm\sqrt{\frac{\left(\sqrt{5}\sqrt{2}\right)^2}{(2+1)^2}-1}\\&=\pm\sqrt{\frac{10}{9}-1}=\pm\sqrt{\frac{1}{9}}\\\tan\theta&=\pm\bf\frac{1}{3}\end{aligned}$