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Help me with q8(iii)

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

Help me with q8(iii)
Help me with q8(iii)

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

A vector which lies in the plane OAB and is also perpendicular to {\bf a} is

\large\text{$\begin{aligned}{\bf u} = 19\,{\bf i}\:-\:4\,{\bf j}\:+\:11\,{\bf k}\end{aligned}$}

 

Penjelasan/Explanation

Vektor/Vectors

Given that \overrightarrow{OA}={\bf a}={\bf i}+2{\bf j}-{\bf k}and\overrightarrow{OB}={\bf b}=3{\bf i}-{\bf j}+2{\bf k}, we want to find a vector which lies in the plane OABand is also perpendicular to{\bf a}.

Let {\bf u}=u_1\,{\bf i}+u_2\,{\bf j}+u_3\,{\bf k} be the vector we want to find. They must satisfy:

  • {\bf a}, {\bf b}, and {\bf u}arecoplanar (Requirement 1).
  • {\bf u}\perp{\bf a} (Requirement 2).

 

Requirement 1

{\bf a}, {\bf b}, and {\bf u} are coplanar if and only if:

\begin{aligned}{\bf u}\cdot({\bf a}\times{\bf b})&=\begin{vmatrix}u_1&u_2&u_3\\ a_1&a_2&a_3\\ b_1&b_2&b_3 \end{vmatrix}=0\end{aligned}

By substituting all components of both {\bf a}and{\bf b}, we get:

\begin{aligned}0&=\begin{vmatrix}u_1&u_2&u_3\\ 1&2&-1\\ 3&-1&2 \end{vmatrix}\\&=4u_1+(-3u_2)+(-u_3)\\&{\quad}-6u_3-u_1-2u_2\\&=3u_1-5u_2-7u_3\\\therefore&\ \ 3u_1-5u_2-7u_3=0\quad...(i)\end{aligned}

 

Requirement 2

{\bf u}\perp{\bf a}if and only if{\bf u}\cdot{\bf a}=0.

So,

\begin{aligned}0&={\bf u}\cdot{\bf a}\\&=\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-1\end{pmatrix}\\&=u_1+2u_2-u_3\\\therefore&\ \ u_1+2u_2-u_3=0\quad...(ii)\end{aligned}

 

Solving (i) and (ii)

From eq. (i), we get:

u_1=-2u_2+u_3\quad...(iii)

\begin{aligned}&(iii)\to(i):\\&3(-2u_2+u_3)-5u_2-7u_3=0\\&\Rightarrow(-6-5)u_2+(3-7)u_3=0\\&\Rightarrow-11u_2-4u_3=0\\&\Rightarrow-11u_2=4u_3\\&\Rightarrow u_2=-\frac{4}{11}u_3\quad...(iv)\end{aligned}

\begin{aligned}(iv)&\to(iii):\\u_1&=-2\left(-\frac{4}{11}u_3\right)+u_3\\&=\left(\frac{8+11}{11}\right)u_3\\&=\frac{19}{11}u_3\end{aligned}

Hence,

\begin{aligned}{\bf u}&=u_1{\bf i}+u_2{\bf j}+u_3{\bf k}\\&=\frac{19u_3}{11}\,{\bf i}\:-\:\frac{4u_3}{11}\,{\bf j}\:+\:u_3\,{\bf k}\end{aligned}

u_3 is "symbolic", means whatever value u_3 equals to, we’ll get a perpendicular vector to {\bf a}. So, let u_3=11, we get:

\begin{aligned}{\bf u}&=\frac{19\cdot11}{11}\,{\bf i}\:-\:\frac{4\cdot11}{11}\,{\bf j}\:+\:11\,{\bf k}\\\\\therefore&\ \ \boxed{\ {\bf u}=19\,{\bf i}\:-\:4\,{\bf j}\:+\:11\,{\bf k}\ }\\&\sf or\\\therefore&\ \ \boxed{\ {\bf u}=\begin{pmatrix}19\\-4\\11\end{pmatrix}\ }\end{aligned}

\blacksquare

 

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Last Update: Sun, 31 Jul 22
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