Kami kembali memberikan contoh Soal Matematika Vektor Kelas XII SMA Part 3 Semoga bermanfaaat bagi adik adik semua yag masih SMA dan juga para Guru matematika semoga apa yang saya tulis semoga bermanfaat.
a) \(frac{\sqrt{2}}{2}\)
b) \(\frac{1}{2\sqrt{2}}\)
c) \(\frac{1}{2\sqrt{2}}\left(0, 1, 1\right)\)
d) \(\frac{1}{2\sqrt{2}}\left(1, 1, 0\right)\)
e) \(\frac{1}{4}\left(1, 1, 0\right)\)
Penyelesaian:
\(\vec{AF}\) = \(\left( 1, 0, 1\right)\)<br>
\(\vec{AP}\) = \(\left( 1, \frac{1}{2}, \frac{1}{2} \right)\)<br>
Maka diperoleh \(\vec{FP}\) = \(\left( 1, \frac{1}{2}, \frac{1}{2} \right)\)<br>
\(\vec{AC}\) = \(\left( 1, 1, 0 \right)\)<br>
Vektor proyeksi \(\vec{FP}\) ke \(\vec{AC}\) adalah<br>
=\(\left|\frac{\vec{FP}\times\vec{AC}}{|\vec{AC}^{2}}\right|\vec{AC}\)= \(\left|\frac{0+frac{1}{2}+0}{(\sqrt{1+1+0})^{2}}\right|\vec{AC}\)<br>
=\(\frac{1}{4}\vec{AC}\) = \(\frac{1}{4}(1, 1, 0)\)
32. Diketahu vektor \(\vec{u}\) = \(\left(2, -1, 1\right)\) dan \(\vec{v}\) = \(\left(-1, 1, -1\right)\). Vektor \(\vec{w}\) panjangnya \(1\) , tegak lurus pada \(\vec{u}\) dan tegak lurus pada \(\vec{v}\) adalah...
a) \(\left(0, 0, 1\right)\)
b) \(\left(0, \frac{1}{2}\sqrt{2}, \frac{1}{2}\sqrt{2}\right)\)
c) \(\left(0, - \frac{1}{2}\sqrt{2}, \frac{1}{2}\sqrt{2}\right)\)
d) \(\left(-\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right)\)
e) \(\left(\frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right)\)
Penyelesaian:
Misal vektor \(\vec{w}\) = \(\left(x, y, z\right)\)<br>
\(\vec{w}\) tegak lurus \(\vec{u}\), maka: \(\vec{w} \times \vec{u}\) = \(0\) \(\Rightarrow\) \(2x-y+z\) = \(0\) ...\(\left(i\right)\)<br>
\(\vec{w}\) tegak lurus \(\vec{v}\), maka: \(\vec{w} \times \vec{v}\) = \(0\) \(\Rightarrow\) \(-x+y-z\) = \(0\) ...\(\left(ii\right)\)<br>
Dari \(\left(i\right)\) dan \(\left(ii\right)\), dengan metode eliminasi:<br>
\(2x - y + z\) = \(0\)<br>
\(-x+y-z \)= \(0\)<br>
Jika di jumlahkan maka \(x \) = \(0\)<br>
dan diperoleh \(y\) = \(z\)<br>
sehingga vektor \(\vec{w}\) = \(\left(0, y, y\right)\)<br>
karena panjang vektor \(\vec{w}\) = \(1\) maka:<br>
\(\Rightarrow\) \(\sqrt{0+y^{2}+y^{2}}\) = \(1\)<br>
\(\Rightarrow\) \(2y^{2}\) = \(1\) \(\Rightarrow\) \(y\) = \(\frac{1}{2}\sqrt{2}\)<br>
Jadi, vektor \(\vec{w}\) = \(\left(0, \frac{1}{2}\sqrt{2}, \frac{1}{2}\sqrt{2}\right)\)
33. Jika vektor tak nol \(\vec{a}\) dan \(\vec{b}\) memenuhi \(\left|\vec{a}+\vec{b}\right|\) = \(\left|\vec{a}-\vec{b}\right|\) maka vektor \(\vec{a}\) dan \(\vec{b}\) ....
a) Membentuk sudut \(90^{\circ}\)
b) \(\left(0, \frac{1}{2}\sqrt{2}, \frac{1}{2}\sqrt{2}\right)\)
c) \(\left(0, - \frac{1}{2}\sqrt{2}, \frac{1}{2}\sqrt{2}\right)\)
d) \(\left(-\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right)\)
e) Berlawanan arah
Penyelesaian:\(\vec{a}\) tegak lurus \(\vec{b}\) = \(0\)
34. Diketahui \(\left|\vec{a}\right|\) = \(4\sqrt{3}\); \(\left|\vec{b}\right|\) = \(5\) ; dan \(\left(\vec{a}+\vec{b}\right)\left(\vec{a}+\vec{b}\right)\) = \(13\). Sudut vektor \(\vec{a}\) dan \(\vec{b}\) adalah...
a) \(150^{\circ}\)
b) \(135^{\circ}\)
c) Membentuk sudut \(45^{\circ}\)
d) Searah
e) \(30^{\circ}\)
Penyelesaian:
Misal, sudut antara vektor \(\left|\vec{a}\right|\) dan \(\left|\vec{b}\right|\) adalah \(\alpha\)<br>
Diketahui<br>
\(\left|\vec{a}\right|\) = \(4\sqrt{3}\); \(\left|\vec{b}\right|\) = \(5\) , maka:<br>
\(\left|\vec{a} + \vec{b}\right|\) =\(\sqrt{\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2} + 2\left|\vec{a}\right|\left|\vec{b}\right|\cos \alpha}\)<br>
=\(\sqrt{\left(4\sqrt{3}\right)^{2}+\left(5\right)^{2}+2\times\left(4\sqrt{3}\right)\left(5\right) \cos \alpha}\)<br>
=\(\sqrt{73+40\sqrt{3}\times \cos \alpha}\)<br>
\(\Rightarrow\) \(\sqrt{\left(\vec{a}+\vec{b}\right)^{2}}\) = \(\sqrt{73+40\sqrt{3}\times \cos \alpha}\)<br>
\(\Rightarrow\) \(\sqrt{13}\) = \(\sqrt{73+40\sqrt{3}\times \cos \alpha}\)<br>
\(\Rightarrow\) \(13\) = \(73 + 40\sqrt{3} \times \cos \alpha\)<br>
\(\Rightarrow\) \(60\) = \(40\sqrt{3} \times \cos \alpha\)<br>
\(\Rightarrow\) \(\frac{-60}{40\sqrt{3}}\) = \(\cos \alpha\) \(\Rightarrow\) \(\cos \alpha\) = \(-\frac{1}{2}\sqrt{3}\)<br>
\(\alpha\) = \(150^{\circ}\) atau \(210^{\circ}\)
35. Agar vektor \(\vec{a}\) = \(\left(x, 4, 7\right)\) dan \(\vec{b}\) = \(\left(6, y, 14\right)\) segaris , haruslah nilai \(x-y\) sama dengan...
a) \(-5\)
b) \(-2\)
c) \(3\)
d) \(4\)
e) \(6\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}\) = \(\left(x, 4, 7\right)\) dan \(\vec{b}\) = \(\left(6, y, 14\right)\)<br>
Jika \(\vec{a}\) dan \(\vec{b}\) segaris maka berlaku \(\vec{a}\) = k\(\vec{b}\)<br>
\(\Rightarrow\) \(\begin{pmatrix} x \\ 4 \\ 7 \end{pmatrix}\) = \(k\begin{pmatrix} 6 \\ y \\ 14 \end{pmatrix}\)<br>
Dari persamaan diatas,diperoleh :<br>
\(7\) = \(k\) \(\times\) \(14\) maka \(k\) = \(\frac{1}{2}\)<br>
Sehingga, \(x\) = \(\frac{1}{2} \times 6\) = \(3\) dan \(4\) = \(\frac{1}{2}\times y\) \(\Rightarrow\) \(y\) = \(8\)<br>
Jadi, nilai dari \(x - y\) = \(3 - 8\) = \(-5\)
36. Diberikan matriks dan vektor - vektor sebagai berikut:<br>
\(A\) = \(\begin{pmatrix} 2 & 2 \\ 2 & -1 \\ 3 & 2 \end {pmatrix}\) ; \(\vec{a}\) = \(\begin{pmatrix} 1 \\ -2 \\2 \end{pmatrix}\) : \(\vec{b}\) = \(\begin{pmatrix} p \\ q \end{pmatrix}\)<br>
\(A^{T}\) menyatakan transpose dari \(A\). jika vektor \(A^{T}\).\(\vec{a}\) tegak lurus dengan \(\vec{b}\) maka \(p\) sama dengan ...
a) \(q\)
b) \(-q\)
c) \(2q\)
d) \(-2q\)
e) \(3q\)
Penyelesaian:
Diketahui:<br>
Matriks \(A\) = \(\begin{pmatrix} 2&2 \\ 2&-1 \\ 3&2 \end{pmatrix}\)<br> \(\Rightarrow\) \(A^{T}\) = \(\begin{pmatrix} 2&2&3 \\ 2&-1&2 \end{pmatrix}\)<br>
\(A^{T} \times \vec{a}\) = \(\begin{pmatrix} 2&2&3 \\ 2&-1&2 \end{pmatrix} \times \begin{pmatrix} 2&2 \\ 2&-1 \\ 3&2 \end{pmatrix}\) = \(\begin{pmatrix}2-4+6 \\ 2+2+4\end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 8 \end{pmatrix}\)<br>
Karena vaktor \(A^{T} \times \vec{a}\) tegak lurus \(\vec{b}\) \(\Rightarrow\) \(A^{T} \times \vec{a} \times \vec{b}\) = \(0\)<br>
Sehingga \(4p+8q\) = \(0\) \(\Rightarrow\) \(p+2q\) = \(0\) \(\Rightarrow\) \(p\) = \(-2q\)
37. Jika \(\vec{p}\), \(\vec{q}\), dan \(\vec{s}\) berturut-turut adalah vektor posisi titik - titik sudut jajaran genjang \(PQRS\) dengan \(PQ\) sejajar \(SR\) maka \(\vec{s}\) =....
a) \(-\vec{p}+\vec{q}+\vec{r}\)
b) \(-\vec{p}-\vec{q}+\vec{r}\)
c) \(\vec{p}-\vec{q}+\vec{r}\)
d) \(\vec{p}-\vec{q}-\vec{r}\)
e) \(\vec{p}+\vec{q}+\vec{r}\)
Penyelesaian:
\(PQRS\) jajargenjang denga \(PQ\) sejajar \(SR\) maka berlaku:<br>
\(PQ\) = \(SR\)<br>
\(\vec{q} - \vec{p}\) = \(\vec{s} - \vec{r}\)<br>
\(\vec{s}\) = \(-\vec{p}+\vec{q}+\vec{r}\)
38. Jika proyeksi vektor \(\vec{u}\) = \(3\vec{i} + 4\vec{j}\) ke vektor \(\vec{v}\) = \(-4\vec{i} + 8\vec{j}\) adalah vektor \(\vec{w}\) maka \(\left|\vec{w}\right|\) adalah...
a) \(\sqrt{5}\)
b) \(5\)
c) \(\sqrt{3}\)
d) \(\vec{p}-\vec{q}-\vec{r}\)
e) \(1\)
Penyelesaian:
Jika proyeksi vektor \(\vec{u}\) = \(3\vec{i} + 4\vec{j}\) ke vektor \(\vec{v}\) = \(-4\vec{i} + 8\vec{j}\) adalah vektor \(\vec{w}\) maka \(\left|\vec{w}\right|\) adalah panjang proyeksi dari vektor \(\vec{u}\) pada \(\vec{v}\).<br>
Sehingga:<br>
\(\left|\vec{w}\right|\)=\(\frac{\vec{u}\times\vec{u}}{\left(\vec{v}\right)}\)=\(\frac{-12+32}{\sqrt{16+64}}\) = \(\frac{20}{\sqrt{80}}\)<br>
=\(\sqrt{\frac{20}{4\sqrt{5}}}\)=\(\sqrt{5}\)
39. Diketahui vektor \(\vec{a}\) = \(\begin{pmatrix} -2 \\ 3 \\ 4 \end{pmatrix}\) dan \(\vec{b}\) = \(\begin{pmatrix} x \\ 0 \\ 3 \end{pmatrix}\). Jika proyeksi vektor \(\vec{a}\) pada \(\vec{b}\) adalah \(\frac{4}{5}\) maka salah satu nilai \(x\) adalah...
a) \(6\)
b) \(4\)
c) \(-4\vec{i}+4\vec{j}-2\vec{k}\)
d) \(-6\vec{i}+4\vec{j}+16\vec{k}\)
e) \(-6\)
Penyelesaian:
\(\vec{a}\) = \(\begin{pmatrix} -2 \\ 3 \\ 4 \end{pmatrix}\) dan \(\vec{b}\) = \(\begin{pmatrix} x \\ 0 \\ 3 \end{pmatrix}\)
maka \(x\) = \(4\)
40. Diketahui vektoe - vektor \(\vec{a}\)=\(2\vec{i}-4\vec{j}+3\vec{k}\), \(\vec{b}\)=\(x\vec{i}+z\vec{j}+4\vec{k}\) , \(\vec{c}\)=\(5\vec{i}-3\vec{j}+2\vec{k}\), dan \(\vec{d}\)=\(2\vec{i}+z\vec{j}+x\vec{k}\), Jika vektor \(\vec{a}\) tegak lurus dengan vektor \(\vec{d}\) maka \(\vec{a} - \vec{b}\) adalah ...
a) \(-6\vec{j}-\vec{k}\)
b) \(-6\vec{i}+4\vec{j}-16\vec{k}\)
c) \(-4\vec{i}+4\vec{j}-2\vec{k}\)
d) \(-6\vec{i}+4\vec{j}+16\vec{k}\)
e) \(4\vec{i}-6\vec{j}-\vec{k}\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}\)=\(2\vec{i}-4\vec{j}+3\vec{k}\), \(\vec{b}\)=\(x\vec{i}+z\vec{j}+4\vec{k}\) , \(\vec{c}\)=\(5\vec{i}-3\vec{j}+2\vec{k}\), dan \(\vec{d}\)=\(2\vec{i}+z\vec{j}+x\vec{k}\)<br>
Vektor \(\vec{a}\) tegak lurus \(\vec{b}\) \(\Rightarrow\) \(\vec{a}\times\vec{b}\) = \(0\)<br>
\(\Rightarrow\) \(2x - 4z + 12\) = \(0\) \(\Rightarrow\) \(2x - 4z \) = \(-12\) ... \(\left(i\right)\)<br>
Vektor \(\vec{c}\) tegak lurus \(\vec{d}\) \(\Rightarrow\) \(\vec{c} \times \vec{d}\) = \(0\)<br>
\(\Rightarrow\) \(10 - 3z + 2x \) = \(0\) \(\Rightarrow\) \(2x - 3z\) = \(-10\)....\(\left(ii\right)\)<br>
Dari \(\left(i\right)\) dan \(\left(ii\right)\), dengan ,etode eliminasi:<br>
\(2x - 4z\) = \(-12\)<br>
\(2x - 3z \)= \(-10\)<br>
Jika dikurangi maka \(z\)=\(-2\) \(\Rightarrow\) \(z\)=\(2\)<br>
Maka, \(x\) = \(-2\)<br>
Sehingga, vektor \(\vec{b}\)=\(-2\vec{i}+2\vec{j}+4\vec{k}\)<br>
Jadi, \(\vec{a} - \vec{b}\) = \(\left(2-\left(-2\right)\right)\vec{i} + \left(-4-2\right)\vec{j} + \left(3-4\right)\vec{k}\)<br>
= \(4\vec{i}-6\vec{j}-\vec{k}\)
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