HOTS Zone : Vektor

Table of Contents

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

Tipe:

StandarSNBTHOTS

No.

P, Q, dan R berturut-turut adalah titik tengah segitiga ABC, seperti gambar di bawah ini. Jika O sembarang titik di luar segitiga ABC, apakah berlaku juga hubungan {\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR}}?

Grup Matematika Zone Id

ALTERNATIF PENYELESAIAN

{\overrightarrow{OP}=\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OB}}
{\overrightarrow{OQ}=\dfrac12\overrightarrow{OB}+\dfrac12\overrightarrow{OC}}
{\overrightarrow{OR}=\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OC}} \begin{aligned} \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}&=\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OB}+\dfrac12\overrightarrow{OB}+\dfrac12\overrightarrow{OC}+\dfrac12\overrightarrow{OC}\\ &=\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OB}+\dfrac12\overrightarrow{OB}+\dfrac12\overrightarrow{OC}+\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OC}\\ &=\left(\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OB}\right)+\left(\dfrac12\overrightarrow{OB}+\dfrac12\overrightarrow{OC}\right)+\left(\dfrac12\overrightarrow{OA}+\dfrac12\overrightarrow{OC}\right)\\ &=\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR} \end{aligned}

Jadi, berlaku hubungan {\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OP}+\overrightarrow{OQ}+\overrightarrow{OR}}.

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No.

Jika beberapa vektor dari $3\vec{i}+3\vec{j}+\sqrt3\vec{k}$, $\vec{i}+\vec{k}$, $\sqrt3\vec{i}+\sqrt{3}\vec{j}+\lambda\vec{k}$ adalah koplanar, maka berapakah nilai λ

1. −1
2. $-\sqrt3$

3. 1
4. $\sqrt3$

5. 3

KPATN 2024 Jenjang Guru

ALTERNATIF PENYELESAIAN

\(\begin{aligned} \begin{vmatrix}3&3&\sqrt3\\1&0&1\\\sqrt3&\sqrt3&\lambda\end{vmatrix}&=0\\[4pt] 3\begin{vmatrix}0&1\\\sqrt3&\lambda\end{vmatrix}-3\begin{vmatrix}1&1\\\sqrt3&\lambda\end{vmatrix}+\sqrt3\begin{vmatrix}1&0\\\sqrt3&\sqrt3\end{vmatrix}&=0\\ 3\left(-\sqrt3\right)-3\left(\lambda-\sqrt3\right)+\sqrt3(\sqrt3)&=0\\ -3\sqrt3-3\lambda+3\sqrt3+3&=0\\ \lambda&=\color{blue}\boxed{\boxed{\color{black}1}} \end{aligned}\)

Jadi, λ = 1.
JAWAB: C

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