Kami kembali memberikan contoh Soal Matematika Vektor Kelas XII SMA Part 2 Semoga bermanfaaat bagi adik adik semua yag masih SMA dan juga para Guru matematika semoga apa yang saya tulis semoga bermanfaat.
a) \(5\sqrt{6}\)
b) \(\frac{5}{6}\sqrt{6}\)
c) \(\frac{6}{5}\sqrt{6}\)
d) \(\frac{5}{6}\sqrt{5}\)
e) \(6\sqrt{5}\)
Penyelesaian:
Diketahu:<br>
\(\vec{u}\) = \(4\vec{i} + 2\vec{j}+ 3\vec{k}\) dan \(\vec{v}\) = \(\vec{i} + \vec{j}+ 2\vec{k}\)<br>
panjang proyeksi vektor \(a\) \(\vec{u}\) pada vektor \(\vec{v}\) sama dengan \(10\) maka:<br>
\(\frac{a\vec{u}\times\vec{v}}{\left|v\right|}\) = \(10\)<br>
\(\Rightarrow\) \(\frac{\left(4a, 2a, 3a\right)\times\left(1, 1, 2\right)}{\sqrt{1^{2}+1^{2}+2^{2}}}\) = \(10\)<br>
\(\Rightarrow\) \(\frac{4a+2a+6a}{\sqrt{6}}\) = \(10\)<br>
\(\Rightarrow\) \(\frac{12a}{\sqrt{6}}\) = \(10\) \(\Rightarrow\) \(12a\) = \(10 \sqrt{6}\)<br>
\(\Rightarrow\) \(a\) = \(\frac{10}{12}\sqrt{6}\) = \(\frac{5}{6}\sqrt{6}\)
12. Diketahui segitiga \(ABC\). Titik \(P\) ditengah \(AC\), dan \(Q\) pada \(BC\) sehingga \(BQ\) = \(QC\).
Jika \(\vec{AB}\) = \(\vec{c}\),\(\vec{AC}\) = \(\vec{b}\), dan \(\vec{BC}\) = \(\vec{a}\) maka \(\vec{PQ}\) =
a) \(\frac{1}{2}\left(-\vec{a}+\vec{b}\right)\)
b) \(\frac{1}{2}\left(\vec{a}+\vec{b}\right)\)
c) \(\frac{1}{2}\left(-\vec{a}+\vec{c}\right)\)
d) \(\frac{1}{2}\left(-\vec{b}+\vec{c}\right)\)
e) \(\frac{1}{2}\left(\vec{b}+\vec{c}\right)\)
Penyelesaian:
\(\vec{PQ}\) = \(\vec{AQ} - \vec{AP}\) <br>
= \(\left(\vec{c} + \frac{1}{2}\vec{a}\right) - \left(\frac{1}{2}\vec{b}\)<br>
= \(\left(\vec{b}-\vec{a}\right) + \frac{1}{2}\vec{a}\right) - \left(\frac{1}{2}\vec{b}\right)\)<br>
= \(\vec{b} - \frac{1}{2}\vec{a} - \frac{1}{2}\vec{b}\)<br>
= \(-\frac{1}{2}\vec{a} + \frac{1}{2}\vec{b}\) = \(\frac{1}{2}\left(-\vec{a}+\vec{b}\right)\)
13. Diketahui vektor \(\vec{a}\) = \(4\vec{i} - 2\vec{j}+ 2\vec{k}\) dan vektor \(\vec{b}\) = \(2\vec{i} - 6\vec{j}+ 4\vec{k}\). Proyeksi vektor orthogonal vektor \(\vec{a}\) dan \(\vec{b}\) adalah
a) \(\vec{i} - \vec{j}+ \vec{k}\)
b) \(\vec{i} - 3\vec{j}+ 2\vec{k}\)
c) \(\frac{1}{2}\left(-\vec{a}+\vec{c}\right)\)
d) \(\frac{1}{2}\left(-\vec{b}+\vec{c}\right)\)
e) \(6\vec{i} - 8\vec{j}+ 6\vec{k}\)
Penyelesaian:
Diketahu:<br>
\(\vec{a}\) = \(4\vec{i} - 2\vec{j}+ 2\vec{k}\) dan vektor \(\vec{b}\) = \(2\vec{i} - 6\vec{j}+ 4\vec{k}\)<br>
Sehingga, Proyeksi vektor orthogonal vektor \(\vec{a}\) dan \(\vec{b}\)<br>
=\(\frac{\vec{a}\times \vec{b}}{|\vec{b}|^{2}}\times \vec{b}\) = \(\frac{4\times2+(-2)(-6)+2\times4}{|\sqrt{2^{2}+(-6)^{2}+4^{2}}|^{2}}\times(2\vec{i}-6\vec{j}+4\vec{k}\)<br>
=\(\frac{8+12+8}{|\sqrt{4+36+16}|^{2}}\times(2\vec{i}-6\vec{j}+4\vec{k}\)<br>
=\(\frac{28}{56}\times (2\vec{i}-6\vec{j}+4\vec{k}\)=\(\vec{i} - 3\vec{j}+ 2\vec{k}\)
14. Nilai \(p\) agar vektor \(pi + 2j - 6k\) dan \(4i - 2j + k\) saling tegak lurus adalah
a) \(6\)
b) \(3\)
c) \(1\)
d) \(-1\)
e) \(-6\)
Penyelesaian:
Dua vektor \(\vec{a}\)dan \(\vec{b}\)saling tegak lurus jika \(\vec{a}\times \vec{b}\)= \(0\)<br>
Sehingga vektor \(i + 2j - 6k\) dan 4i - 3j + k saling tegak lurus jika:<br>
\(4p + 2(-3) + (-6) \times 1 \) = \(0\)<br>
\(\Rightarrow\) \(4p-6-6\) =\(0\)<br>
\(\Rightarrow\) \(4p\) = \(12\) \(\Rightarrow\) \(p\) =
\(3\)
15. Diketahui vektor \(\vec{u}\) = \(3\vec{i} + 2\vec{j} - \vec{k}\) dan \(\vec{v}\) = \(3\vec{i} + 9\vec{j}- 12\vec{k}\). Jika vektor \(2\vec{u} - a\vec{v}\) tegak lurus terhadap \(\vec{v}\) maka nilai \(a\) adalah
a) \(-1\)
b) \(-\frac{1}{3}\)
c) \(1\)
d) \(\frac{1}{3}\)
e) \(3\)
Penyelesaian:
\((6-3a))\times+(4-9a)\times9+(-2+12a)(-12)=0\)<br>
\(18-9a+36-81a+24-144a\)=\(0\)<br>
\(78-234a\) =\(0\)
\(a\)=\(\frac{78}{234}\) =\(\frac{1}{3}\)
16. Jika \(\vec{OA}\) = \(\left(1,2\right)\) ; \(\vec{OB}\) = \(\left(4,2\right)\) dan \(\Theta\) = \(\angle \left(\vec{OA},\vec{OB}\right)\) maka \(\tan \Theta\) =
a) \(\frac{3}{5}\)
b) \(\frac{3}{4}\)
c) \(\frac{4}{3}\)
d) \(\frac{9}{16}\)
e) \(\frac{16}{9}\)
Penyelesaian:
\(\vec{OA}\) = \(\left(1,2\right)\) ; \(\vec{OB}\) = \(\left(4,2\right)\) dan \(\Theta\) = \(\angle \left(\vec{OA},\vec{OB}\right)\)
\(tan\Theta\) =\(\frac{3}{4}\)
17. Diketahui titik \(A\left(5, 1, 3\right)\) ; \(B\left(2, -1, -1\right)\) dan \(C\left(4, 2, -4\right)\). Besar sudut \(ABC\) =
a) \(\pi\)
b) \(\frac{\pi}{2}\)
c) \(\frac{\pi}{3}\)
d) \(\frac{\pi}{6}\)
e) \(0\)
Penyelesaian:
\(A\left(5, 1, 3\right)\) ; \(B\left(2, -1, -1\right)\) dan \(C\left(4, 2, -4\right)\)
\(\cos\alpha\)=\(0\) \(\alpha\) = \(\frac{\Pi}{2}\)
18. Diketahui \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) vektor dalam dimensi-\(3\). Jika \(\vec{a}\) tegak lurus \(\vec{b}\) dan \(\vec{a}\) tegak lurus \(\left(\vec{b} + 2\vec{c}\right)\) maka \(\vec{a}\left(\vec{b}+2\vec{c}\right)\)=
a) \(4\)
b) \(2\)
c) \(1\)
d) \(0\)
e) \(-1\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}, \vec{b}\) dan \(\vec{c}\) vektor dalam dimensi \(3\)<br>
\(\vec{a}\) tegak lurus \(\vec{b}\) \(\Rightarrow\) \(\vec{a} \times \vec{b}\) = \(0\)<br>
\(\vec{a}\) tegak lurus \(\left(\vec{b} + 2\vec{c}\right)\) \(\Rightarrow\) \(\vec{a} \times \left(\vec{b} + 2\vec{c} \right)\) = \(0\)<br>
\(\Rightarrow\) \(\vec{a} \times \vec{b} + \vec{a} \times 2\vec{c}\) = \(0\)<br>
\(\Rightarrow\) \(0 + \vec{a} \times 2\vec{c}\) = \(0\) \(\Rightarrow\) \(\vec{a} \times 2\vec{c}\) = \(0\)<br>
sehingga, nilai dari:<br>
\(\vec{a}\left(\vec{b}+2\vec{c}\right)\) = \(\vec{a} \times \vec{b} - \vec{a} \times 2\vec{c}\)<br>
=\(0-0\)=\(0\)
19. Diketahui titik \(A\left(3, 1, -4\right)\), \(B\left(3, -4, 6\right)\) dan \(C\left(-1, 5, 4\right)\). Titik \(P\) membagi \(AB\) sehingga \(AP : PB\) = \(3 : 2\) maka vektor yang diwakili oleh \(\vec{PC}\) adalah
a) \(\begin{pmatrix} -4 \\ 3 \\ -6 \end{pmatrix}\)
b) \(\begin{pmatrix} -4 \\ 3 \\ 6 \end{pmatrix}\)
c) \(\begin{pmatrix} -4 \\ -7 \\ -2 \end{pmatrix}\)
d) \(\begin{pmatrix} 4 \\ -7 \\ -2 \end{pmatrix}\)
e) \(\begin{pmatrix} -4 \\ 7 \\ 2 \end{pmatrix}\)
Penyelesaian:
Titik \(P\) = \(\frac{3/vec{b} + 2/vec{a}}{3+2}\)<br>
=\(\frac{3\left(3,-4,6\right) + 2\left(3, 1, -4\right)}{5}\)<br>
=\(\frac{\left(9, -12, 18\right) + \left(6, 2, -8\right)}{5}\)<br>
=\(frac{\left(15, -10, 10\right)}{5}\) = \(\left(3, -2, 2\right)\)<br>
Sehingga:<br>
\(\vec{PC}\) = \(\vec{PB} + \vec{BC}\) = \(\left(\vec{b} - \vec{c}\right) + \left(\vec{c} - \vec{b}\right)\)<br>
=\(\left(\begin{bmatrix} 3 \\ -4 \\ -6 \end{bmatrix} - \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix}\right) + \left(\begin{bmatrix} -1 \\ 5 \\ 4 \end{bmatrix} - \begin{bmatrix} 3 \\ -4 \\ 6 \end{bmatrix}\right)\)<br>
=\(\begin{bmatrix} 0 \\ -2 \\ 4 \end{bmatrix} + \begin{bmatrix} -4 \\ 9 \\ -2 \end{bmatrix}\) = \(\begin{bmatrix} 4 \\ -7 \\ -2 \end{bmatrix}\)
20. Diketahui titik \(A\left(1, -2, -8\right)\) dan titik \(B\left(3, -4, 0\right)\). Titik \(P\) terletak pada perpanjangan \(AB\) sehingga \(\vec{AP}\) = \(-3\vec{PB}\).<br>
Jika \(\vec{b}\) merupakan vektor posisi titik \(P\) maka \(\vec{p}\) = .....
a) \(4\vec{i}-5\vec{j}+4\vec{k}\)
b) \(4\vec{i}-5\vec{j}-4\vec{k}\)
c) \(-\vec{j}-12\vec{k}\)
d) \(-3\vec{i}-\vec{j}-12\vec{k}\)
e) \(-\vec{i}-5\vec{j}-2\vec{k}\)
Penyelesaian:
Titik \(P\) = \(\frac{3\vec{b} + 2\vec{a}}{3+2}\)<br>
=\(\frac{3\left(3,-4,6\right) + 2\left(3, 1, -4\right)}{5}\)<br>
=\(\frac{\left(9, -12, 18\right) + \left(6, 2, -8\right)}{5}\)<br>
=\(\frac{\left(15, -10, 10\right)}{5}\) = \(\left(3, -2, 2\right)\)<br>
Sehingga:<br>
\(\vec{PC}\) = \(\vec{PB} + \vec{BC}\) = \(\left(\vec{b} - \vec{c}\right) + \left(\vec{c} - \vec{b}\right)\)<br>
=\(\left(\begin{bmatrix} 3 \\ -4 \\ -6 \end{bmatrix} - \begin{bmatrix} 3 \\ -2 \\ 2 \end{bmatrix}\right) + \left(\begin{bmatrix} -1 \\ 5 \\ 4 \end{bmatrix} - \begin{bmatrix} 3 \\ -4 \\ 6 \end{bmatrix}\right)\)<br>
=\(\begin{bmatrix} 0 \\ -2 \\ 4 \end{bmatrix} + \begin{bmatrix} -4 \\ 9 \\ -2 \end{bmatrix}\) = \(\begin{bmatrix} 4 \\ -7 \\ -2 \end{bmatrix}\)
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