Kami kembali memberikan contoh Soal Matematika Vektor Kelas XII SMA Part 5 Semoga bermanfaaat bagi adik adik semua yag masih SMA dan juga para Guru matematika semoga apa yang saya tulis semoga bermanfaat.
a) \(4\sqrt{19}\)
b) \(\sqrt{19}\)
c) \(4\sqrt{7}\)
d) \(2\sqrt{7}\)
e) \(4\sqrt{7}\)
Penyelesaian:
\(\left|\vec{a}\right|\) , \(\left|\vec{b}\right|\), dan \(\left|\vec{a} - \vec{b} \right|\) berturut - turut adalah \( 4, 6, 2\sqrt{12}\)<br>
\(|\vec{a} + \vec{b} - \vec{b}\) = \(2\sqrt{7}\)
52. Diketahui \(\left|\vec{a}\right|\) = \(\sqrt{2}\), \(\left|\vec{b}\right|\) = \(\sqrt{9}\), \(\left|\vec{a}+\vec{b}\right|\) = \(\sqrt{5}\). Besar sudut antara vektor \(\vec{a}\) dan \(\vec{b}\) adalah...
a) \(45^{\circ}\)
b) \(90^{\circ}\)
c) \(120^{\circ}\)
d) \(135^{\circ}\)
e) \(150^{\circ}\)
Penyelesaian:
Vektor \(a\) dan vektor \(b\) membentuk sudut \(\beta\) sehingga berlaku \(a \times b\) =\(\left|\vec{a}\right| \left|\vec{b}\right| \cos \beta\)<br>
\(\left|\vec{a}+\vec{b}\right|\) = \(\sqrt{5}\)<br>
\(\rightarrow\) \(a \times a + b \times b + 2 \left( a \times b\right)\) =\(5\)<br>
\(\rightarrow\) \( \left|a\right|^{2} + \left|b\right| + 2 \left(\left|a\right| \left|b\right| \cos \beta\right)\) = \(5\)<br>
\(\rightarrow\) \( 2 + 9 + 2 \left(\sqrt{2} \times 3 \times \cos \beta \right)\) = \(5\)<br>
\(\rightarrow\) \(\cos \beta\) = \(\frac{5-11}{6 \sqrt{2}}\) = \(\frac{1}{2}\sqrt{2}\)<br>
\(\rightarrow\) \(\beta\) = \(135^{\circ}\)
53. Diketahui vektor \(\vec{a}\) dan \(\vec{b}\) dengan \(\left|\vec{a}\right|\) = \( 4 \) ; \(\left|\vec{b}\right|\) = \( 3 \), dan \(\left|\vec{a} + \vec{b} \right|\) = \(5\) . Jika \(\Theta\) adalah sudut antara vekotr \(\vec{a}\) dan \(\vec{b}\) , nilai \(\cos 20\) adalah...
a) \(1\)
b) \(\frac{4}{5}\)
c) \(0\)
d) \(-\frac{4}{5}\)
e) \(-1\)
Penyelesaian:
\(\left|\vec{a} + \vec{b} \right|\) = \(5\)<br>
\(\sqrt{\left(\vec{a} + \vec{b} \right)^{2}}\) = \(5\)<br>
\(\vec{a} \times \vec{a} + \vec{b} \times \vec{b} +2\left|\vec{a}\right|\left| \vec{a}\right| \) = \(5^{2}\)<br>
\( \left|\vec{a}\right|^{2} + \left|\vec{b}\right|^{2} + 2 \left|\vec{a}\right| \left|\vec{b}\right| \cos \theta\) = \(25\)<br>
\(4^{3} + 3^{2} + 2 \times 4 \times 3 \times \cos \theta \) = \(25\)<br>
\(24 \cos \theta\) = \(0\)<br>
\(\cos \theta\) = \(0\)<br>
\(\theta\) = \(90^{\circ}\)<br>
jadi , \(\cos 2\theta \) = \(\cos 180^{\circ}\) = \(-1\)
54. Diketahui titik \(A(\left(2, 7, 8\right) \), \(B(\left(1, -1, 1\right) \), dan \(A \left(0,3, 2\right) \). Jika \(\vec{AB}\) wakil \(\vec{u}\) dab \(\vec{BC}\) wakil \(\vec{v}\), maka proyeksi orthogonal vektor \(\vec{u}\) pada \(\vec{v}\) adalah....
a) \(-3\vec{i} - 6\vec{j} - 9\vec{k}\)
b) \(\vec{i} +2b\vec{j} +3 \vec{k}\)
c) \(\frac{1}{3}\vec{i} + \frac{2}{3}\vec{j} +\vec{k}\)
d) \(9\vec{i} - 18\vec{j} - 27\vec{k}\)
e) \(3\vec{i} + 6\vec{j} + 9\vec{k}\)
Penyelesaian:
Vektor \(\vec{u}\) = \(\vec{AB}\) = \(B-A\) = \(\begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 7 \\ 8 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ -6 \\ -9 \end{pmatrix}\)<br>
Vektor \(\vec{v}\) = \(\vec{CB}\) = \(C-B\) = \(\begin{pmatrix} 0 \\ 3 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix}\) = \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)<br>
\(\left|\vec{v}\right|\) = \(\sqrt{\left(1\right)^{2} + 2^{2} + 3^{2}}\) = \(\sqrt{14}\) <br>
proyeksi vektor orthogonal vektor \(\vec{u}\) pada \(\vec{v}\) = \(\frac{\vec{u}\times\vec{v} \times \vec{u}}{\left|\vec{v}\right|^{2}}\)<br>
=\(\frac{\begin{pmatrix} -3 \\ -6 \\ -9 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}}{\left(\sqrt{14}\right)^{2}}\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) = \(\frac{-3-12-27}{14}\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)<br>
=\( -3\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)=\(\begin{pmatrix} -3 \\ -6 \\ -9 \end{pmatrix}\) \(\rightarrow\) \(-3\vec{i} - 6\vec{j} - 9\vec{k}\)
55. Vektor \(z\) adalah proyeksi vektor \(x\) = \(\left(-\sqrt{3}, 3, 1\right)\) pada vektor vektor \(y\) = \(\left(\sqrt{3}, 2, 3\right)\). Panjang vektor \(z\) =
a) \(\frac{1}{2}\)
b) \(1\)
c) \(\frac{3}{2}\)
d) \(2\)
e) \(\frac{5}{2}\)
Penyelesaian:
Panjang vektor :<br>
\(y\) = \(\left|y\right|\) = \(\sqrt{\left(\sqrt{3}\right)^{2} + 2^{2} + 3^{2}}\) = \(4\)<br>
Vektor \(z\) adalah proyeksi \(x\) pada \(y\) sehingga <br>
\(z\) = \(\frac{x \times y}{\left|y\right|^{2}} \times y\) =\(\frac{\begin{pmatrix}-\sqrt{3} \\ 3 \\ 1\end{pmatrix} \times \begin{pmatrix} \sqrt{3} \\ 2 \\ 3\end{pmatrix} }{4^{2}}\begin{pmatrix} \sqrt{3} \\ 2 \\ 3\end{pmatrix}\) <br>
\(\frac{-3 + 6 + 3}{16}\begin{pmatrix} \sqrt{3} \\ 2 \\ 3\end{pmatrix}\) =\( \frac{3}{8}\begin{pmatrix} \sqrt{3} \\ 2 \\ 3\end{pmatrix}\) = \(\begin{pmatrix} \frac{3\sqrt{3}}{8} \\ \frac{3}{4} \\ \frac{9}{8}\end{pmatrix}\)<br>
Panjang vektor \(z\)<br>
\(\left|z\right|\) = \(\sqrt{(3\sqrt{3}}{8})^{2} + (\frac{3}{4})^{2} + (\frac{9}{8})^{2}\)<br>
=\(\sqrt{\frac{27}{64}+\frac{9}{16}+\frac{81}{64}}\)<br>
=\(\sqrt{\frac{27+36+81}{64}}\) = \(\frac{3}{2}\)
56. Diketahui vektor \(\vec{a}\) = \(i - 2j +k \) dan \(b\) \(\vec{b}\) = \(3i + j - 2k)\). Vektor \(\vec{c}\) mewakili vektor hasil proyeksi orthogonal vektor \(\vec{b}\) pada hasil vektor \(\vec{a}\) maka vektor \(\vec{c}\) = ....
a) \(-\frac{1}{6}\left(i-2j+k\right)\)
b) \(-\frac{1}{6}\left(3i-2j+2k\right)\)
c) \(-\frac{1}{14}\left(1 -2j+k\right)\)
d) \(-\frac{1}{14}\left(3i +j+2k\right)\)
e) \(\frac{1}{6}\left(1 -2j+k\right)\)
Penyelesaian:
Panjang vektor \(\vec{a}\)<br>
\(|\vec{a}|\) = \(\sqrt{(1)^{2} +(-2)^{2}+(1)^{2}}\) = \(\sqrt{6}\)<br>
vektor \(\vec{c}\) mewakili vektor hasil proyeksi orthogonal vektor \(\vec{b}\) pada \(\vec{a}\),<br>
\(\vec{c}\) = \(\frac{\vec{b} \times \vec{a}}{|\vec{a}|^{2}} \times \vec{a}\) = \(\frac{\begin{pmatrix} 3 \\ 1 \\ -2\end {pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 1\end {pmatrix} }{\sqrt{6})^{2}}\begin{pmatrix} 1 \\ -2 \\ 1\end {pmatrix}\)<br>
=\(\frac{3-2-2}{6}\begin{pmatrix} 1 \\ -2 \\ 1\end {pmatrix} \)<br>
=\(\frac{-1}{6}\begin{pmatrix} 1 \\ -2 \\ 1\end {pmatrix} \)<br>
=\(-\frac{1}{6}\left(i-2j+k\right)\)
57. Diketahui \(\vec{p}\) = \(\begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}\), \(\vec{q}\) = \(\begin{pmatrix} 1 \\ 2x \\ 2 \end{pmatrix}\), dan proyeksi scalar vektor \(\vec{q}\) pada \(\vec{p}\) adalah \(1\frac{1}{7}\) . Nilai \(x\) =..
a) \(-2\)
b) \(-1\)
c) \(0\)
d) \(1\)
e) \(2\)
Penyelesaian:
\(\left|\vec{p}\right|\) = \(\sqrt{2^{2}+3^{2}+6^{2}}\) = \(\sqrt{49}\) = \(7\)<br>
Panjang proyeksi scalar vektor \(\vec{q}\) pada \(\vec{p}\frac{\vec{q}\times\vec{p}}{|\vec{p}|}\)<br>
\(1\frac{1}{7}\) = \(\frac{}{}\)<br>
\(\frac{\begin{pmatrix} 1 \\ 2x \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}}{7}\) = \(\frac{8}{7}\)=\frac{2+6x+12}{7} \(\rightarrow\) \(x\) = \(-1\)
58. Diketahui vektor \(\vec{u}\) = \(\begin{pmatrix} 3 \\ -1 \\1 \end{pmatrix}\) dan vektor \(\vec{v}\) = \(\begin{pmatrix} 2 \\ p \\2 \end{pmatrix}\). Jika proyeksi scalar orthogonal \(\vec{u}\) pada arah vektor \(\vec{v}\) sama dengan setengah panjang vektor \(\vec{v}\), maka nilai \(p\)=....
a) \(-4\) atau \(-2\)
b) \(-4\) atau \(2\)
c) \(4\) atau \(-2\)
d) \(8\) atau \(-1\)
e) \(-8\) atau \(1\)
Penyelesaian:
Panjang vektor \(\vec{v}\) = \(|\vec{v}| 2 (\vec{u} \times \vec{v})\) = \(|\vec{v}|^{2}\)<br>
Proyeksi skalar orthogonal \(\vec{v}\) pada arah vektor \(\vec{v}\) sama dengan setengah panjang \(\vec{v}\)<br>
\(\frac{\vec{u} \times \vec{v}}{|\vec{v}|}\) = \(\frac{1}{2}|\vec{v}|\)<br>
\(2(\vec{u} \times \vec{v})\) = \(|\vec{v}|^{2}\) = \(\sqrt{2^{2} + p^{2} + 2^{2}}\) = \(\sqrt{8 + p^{2}}\)<br>
\(\Rightarrow\) \(2\left(\begin{pmatrix} 3 \\ -1 \\1 \end{pmatrix} \times \begin{pmatrix} 2 \\ p \\2 \end{pmatrix}\right)\) = \(8 + p^{2}\)<br>
\(\Rightarrow\) \(2(6 - p +2 )\) = \(8 + p^{2}\)<br>
\(\Rightarrow\) \(p^{2}+2p-8\) = \(0\)<br>
\(\Rightarrow\) \(\left(p+4\right)\left(p-2\right)\) = \(0\)<br>
Jadi, \(p\) = \(-4\) atau \(p\) = \(2\)
59. Diketahui vektor \(\vec{a}\) = \(\begin{pmatrix} -2 \\ 3 \\ 4 \end{pmatrix}\) dan \(\begin{pmatrix} x \\ 0 \\ 3 \end{pmatrix}\). Jika panjang proyeksi vektor \(\vec{a}\) pada \(\vec{b}\) adalah \(\frac{4}{5}\), maka salah satu nilai \(x\) adalah....
a) \(6\)
b) \(4\)
c) \(2\)
d) \(-4\)
e) \(-6\)
Penyelesaian:
Panjang vektor \(\vec{b}\),<br>
\(\left|\vec{b}\right|\) = \(\sqrt{x^{2}+0^{2}+3^{2}}\) = \(\sqrt{x^{2} + 9}\)<br>
Panjang proyeksi vektor \(\vec{a}\) dan \(\vec{b}\) = \(\frac{4}{5}\),<br>
maka<br>
\(\frac{\vec{a} \times \vec{a}}{\left|\vec{b}\right|}\) = \(\frac{4}{5}\)<br>
\(\frac{\begin{pmatrix} -2 \\ 3 \\ 4 \end{pmatrix} \times \begin{pmatrix} x \\ 0 \\ 3 \end{pmatrix} }{\sqrt{x^{2}+9}}\) = \(\frac{4}{5}\)<br>
\(\frac{-2x+0+12}{\sqrt{x^{2}+9}}\) = \(\frac{4}{5}\)<br>
\(\frac{-2x+0+12}{\sqrt{x^{2}+9}}\) = \(\frac{4}{5}\)<br>
\(-10x+60\) = \(4\sqrt{x^{2}+9}\)<br>
\(-5x+30\) = \(2\sqrt{x^{2}+9}\)<br>
\(25x^{2} - 300x + 900 \) = \(4x^{2} + 36\)<br>
\(21x^{2} - 300x + 864\) = \(0\)<br>
\(7x^{2} - 100x + 288\) = \(0\)<br>
\(\left(7x-72\right)\left(x-4\right)\) = \(0\)<br>
\(x\) = \(\frac{72}{7}\) atau \(x\) = \(4\)
60. Diketahui vektor \(\vec{a}\) = \(4\vec{i} - 2\vec{j} + 2\vec{k}\), dan \(\vec{b}\) = \(2\vec{i} - 6\vec{j} + 4\vec{k}\). Proyeksi vektor orthogonal vektor \(\vec{a}\) pada vektor \(\vec{b}\) adalah...
a) \(\vec{i} - \vec{j} + \vec{k}\)
b) \(\vec{i} -3 \vec{j} +2 \vec{k}\)
c) \(\vec{i} -4 \vec{j} +4 \vec{k}\)
d) \(2\vec{i} - j\vec{j} + \vec{k}\)
e) \(6\vec{i} - 8\vec{j} + 6\vec{k}\)
Penyelesaian:
\(|\vec{b}\) = \(\sqrt{2^{2}+ \left(-6\right)^{2} + 4^{2}}\) = \(\sqrt{56}\)<br>
proyeksi orthogonal vektor \(\vec{a}\) pada vektor \(\vec{b}\)<br>
=\(\frac{8+12+8}{56}\begin{pmatrix} 2 \\ -6 \\ 4\end{pmatrix}\)<br>
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