Kami kembali memberikan contoh Soal Matematika Vektor Kelas XII SMA Part 1 Semoga bermanfaaat bagi adik adik semua yag masih SMA dan juga para Guru matematika semoga apa yang saya tulis semoga bermanfaat.
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) \(2\vec{i}- \vec{j}+ \vec{k}\)
e) \(6\vec{i}- 8\vec{j}+ 6\vec{k}\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}\)=\(4\vec{i}- 2\vec{j}+ 2\vec{k}\) dan \(\vec{b}\)=\(2\vec{i}- 6\vec{j}+ 4\vec{k}\)<br>
Sehingga, proyeksi vektor orthgonal \(\vec{a}\) pada \(\vec{b}\)<br>
=\(\frac{4\times2+(-2)(-6)+2\times4}{\left | \sqrt{2^{2}+(6)^{2}+4^{2}} \right |}\times \left ( 2\vec{i}-6\vec{j}+4\vec{k} \right ) )\)<br>
=\(\frac{8+12+8}{56}\times \left ( 2\vec{i}-6\vec{j}+4\vec{k} \right )\)<br>
=\(\frac{28}{56}\times\left ( 2\vec{i}-6\vec{j}+4\vec{k} \right )\)<br>
=\(\vec{i}- 3\vec{j}+ 2\vec{k}\)
2. Diketahui vektor – vektor \(\vec{u}\)=\(\vec{i}+ \sqrt{2}\vec{j}+ \sqrt{5}\vec{k}\) ; \(\vec{v}\)=\(\vec{i}- \sqrt{2}\vec{j}+ \sqrt{5}\vec{k}\) <br>
Sudut antara vektor \(\vec{u}\) dan \(\vec{v}\) adalah
a) \(30^{\circ}\)
b) \(45^{\circ}\)
c) \(60^{\circ}\)
d) \(90^{\circ}\)
e) \(120^{\circ}\)
Penyelesaian:
Diketahui:<br>
\(\vec{u}\)=\(\vec{i}+ \sqrt{2}\vec{j}+ \sqrt{5}\vec{k}\) ; \(\vec{v}\)=\(\vec{i}- \sqrt{2}\vec{j}+ \sqrt{5}\vec{k}\)<br>
misal, sudut antara \(\vec{u}\) dan \(\vec{v}\) adalah \(\alpha \) maka:<br>
\(1-2+5\) = \(\sqrt{1+2+5}\times \sqrt{1+2+5}\cos \alpha \)<br>
\(4\)=\(8 \cos \alpha\)<br>
\(\frac{1}{2}\)=\(\cos \alpha\)<br>
\(\alpha\) = \(60^{\circ}\)
3. Diketahui titik \(A(2, 7, 8)\); \(B(-1, 1, -1)\); \(C(0, 3, 2)\). Jika \(\vec{AB}\) wakil \(\vec{u}\) dan \(\vec{BC}\) wakil \(\vec{v}\) maka proyeksi orthogonal vektor \(\vec{u}\) dan \(\vec{v}\) adalah
a) \(-3\vec{i}- 6\vec{j}- 9\vec{k}\)
b) \(\vec{i}+ 2\vec{j}+ 3\vec{k}\)
c) \(\frac{1}{3}\vec{i}+ \vec{j}+\vec{k}\)
d) \(-9\vec{i}- 18\vec{j}- 27\vec{k}\)
e) \(3\vec{i}+ 6\vec{j}+ 9\vec{k}\)
Penyelesaian:
Diketahui:<br>
\(A(2, 7, 8)\); \(B(-1, 1, -1)\); \(C(0, 3, 2)\)<br>
Sehingga:<br>
\(\vec{u}\)=\(\vec{AB}\)=\(\left ( \left ( -1-2 \right )\vec{i}+\left ( 1-7 \right )\vec{j}+\left ( -1-8 \right )\vec{k} \right )\)=\(\left (-3\vec{i}-6\vec{6}-9\vec{k} \right )\)<br>
\(\vec{v}\)=\(\vec{BC}\)=\(\left ( \left ( 0-(-1) \right )\vec{i}+\left ( 3-1 \right )\vec{j}+\left ( 2-(1)\vec{k} \right ) \right )\)= \(\left ( \vec{i}+2\vec{j}+3\vec{k} \right )\)
Maka, proyeksi ortogonal vektor \(\vec{u}\) dan \(\vec{v}\)<br>
=\(\frac{\vec{u}\vec{v}}{\left | \vec{v} \right |^{2}}\times \vec{v}\)
=\(\frac{-3-12-27}{\left | \sqrt{1+4+9} \right |^{2}}\times \left ( \vec{i}+2\vec{j}+3\vec{k} \right )\)<br>
=\(\frac{-42}{14}\times \left ( \vec{i}+2\vec{j}+3\vec{k} \right )\)<br>
=\(-3 \left ( \vec{i}+2\vec{j}+3\vec{k} \right )\)<br>
=\(-3\vec{i}- 6\vec{j}- 9\vec{k}\)
4. Diketahui vektor \(\vec{a}\)=\(\vec{i}- x\vec{j}+ 3\vec{k}\) ; \(\vec{b}\)=\(2\vec{i}+ \vec{j}- \vec{k}\) ; \(\vec{c}\)=\(\vec{i}+ 3\vec{j}+ 2\vec{k}\). Jika \(\vec{a}\) tegak lurus \(\vec{b}\) maka \(2\vec{a}(\vec{b} - \vec{c})\) adalah
a) \(-20\)
b) \(-12\)
c) \(-10\)
d) \(-8\)
e) \(-1\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}\)=\(\vec{i}- x\vec{j}+ 3\vec{k}\)<br>
\(\vec{b}\)=\(2\vec{i}+ \vec{j}- \vec{k}\)<br>
\(\vec{c}\)=\(\vec{i}+ 3\vec{j}+ 2\vec{k}\)<br>
karena \(\vec{a}\) tegak lurus \(\vec{b}\) maka \(\vec{a}\times\vec{b}\)=\(0\)<br>
\(\rightarrow\) \(1 \times 2 + \left(-x\right) 1 + 3\left(-1\right) \)= \(0\)<br>
\(\rightarrow\) \(2 - x 3\) = \(0\) \(\rightarrow\) \(x\) = \(-1\)<br>
Jadi, nilai dari \(2\vec{a}(\vec{b} - \vec{c})\)<br>
\(=\)\(2\left(\vec{i}- x\vec{j}+ 3\vec{k} \right)\left(\left(2-1\right)\vec{i}+\left(1-3\right)\vec{j}+\left(-1-2\right)\vec{k}\right)\)<br>
\(=\)\(\left(2\vec{i}+ 2\vec{j}+ 6\vec{k}\right)\left(\vec{i}- 2\vec{j}- 3\vec{k}\right)\)<br>
\(=\)\(2-4-18\)=\(20\)
5. Pada persegi panjang \(OACB\), \(D\) adalah titik tengah \(OA\) dan \(P\) titik potong \(CD\) dengan diagonal \(AB\). Jika \(\vec{a}\)=\(\vec{OA}\) dan \(\vec{b}\)=\(\vec{OC}\) maka \(\vec{CP}\) =
a) \(\frac{1}{3}\vec{a}+\frac{2}{3}\vec{b}\)
b) \(\frac{1}{3}\vec{a}-\frac{2}{3}\vec{b}\)
c) \(-\frac{1}{3}\vec{a}-\frac{2}{3}\vec{b}\)
d) \(-\frac{2}{3}\vec{a}-\frac{1}{3}\vec{b}\)
e) \(-\frac{2}{3}\vec{a}+\frac{1}{3}\vec{b}\)
Penyelesaian:
\(CP : DP\) = \( 2 : 1\)<br>
Maka, \(\vec{CP}\) = \(\frac{2}{3}\vec{CD}\)=\(\left(\vec{CA}+\vec{AD}\right)\)<br>
\(\vec{CP}\)=\(\frac{2}{3}\left(\vec{a}-\vec{b}-\frac{1}{2}\vec{a}\right)\) = \(\frac{1}{3}\vec{a}-\frac{2}{3}\vec{b}\)
6. Diketahui vektor \(\vec{u}\)= \(\left(a, -2, -1\right)\) dan \(\vec{v}\) = \(\left(a, a, -1\right)\). Jika vektor \(\vec{u}\) tegak lurus \(\vec{v}\) maka nilai \(a\) adalah
a) \(-1\)
b) \(0\)
c) \(1\)
d) \(2\)
e) \(3\)
Penyelesaian:
Diketahui:<br>
Vektor \(\vec{u}\)= \(\left(a, -2, -1\right)\) dan \(\vec{v}\) = \(\left(a, a, -1\right)\)<br>
vektor \(\vec{u}\) \(\bot\) \(\vec{v}\) maka<br>
\(\vec{u} \times \vec{v}\) = \(0\)<br>
\(\rightarrow\) \(a \times a + \left(-2 \right) \times a + \left(-1\right)\left(-1\right)\) = \(0\)<br>
\(\rightarrow\) \(a^{2} - 2a + 1\) = \(0\)<br>
\(\rightarrow\) \(\left(a-1\right)^{2}\) =\(0\) \(\rightarrow\) \(a\) = \(1\)
7. Diketahui titik \(P\left(2, 7, 8\right)\) dan \(Q\left(-1, 1, -1\right)\). Titik \(R\) membagi \(PQ\) di dalam dengan perbandingan \(2 : 1\), panjang \vec{PR} =
a) \(\sqrt{4}\)
b) \(\sqrt{6}\)
c) \(\sqrt{12}\)
d) \(\sqrt{14}\)
e) \(\sqrt{56}\)
Penyelesaian:
Vektor \(R\) = \frac{2\left(-1, 1, -1\right) + 1\left(2, 7, 8 \right)}{2+1}<br>
=\(\frac{\left(-2, 2, -2\right)+\left(2, 7, 8\right)}{3}\) = \(\left(\frac{0. 9, 6}{3}\right)\)<br>
=\(\left(0, 3, 2 \right)\)<br>
Maka, \vec{PR} = \(\left(2-0, 7-3, 8-2 \right)\)<br>
=\(\left(2, 4, 6 \right)\)<br>
Jadi, panjang \(\vec{PR}\)=\(\left|\vec{PR}\right|\)<br>
=\(\sqrt{2^{2}+4^{2}+6^{2}}\)=\(\sqrt{56}\)
8. Diketahui vektor \(\vec{u}\) = \(\left(a^{3}, 3, 4a\right)\) dan \(\vec{v}\) =\(\left(2, -7a^{2}, 9\right)\) dengan \(0 < a < 8\). Nilai mksimum \(\vec{u} \times \vec{v}\) adalah
a) \(108\)
b) \(17\)
c) \(15\)
d) \(6\)
e) \(1\)
Penyelesaian:
Diketahui:<br>
vektor \(\vec{u}\) = \(\left(a^{3}, 3, 4a\right)\) dan \(\vec{v}\) =\(\left(2, -7a^{2}\right)\)<br>
Sehingga \(\vec{u}\) \times \(\vec{v}\) = \(2a^{2} - 21a^{2} + 36a\)<br>
Nilai maksimum dari\(\vec{u} \times \vec{v}\) diperoleh ketika \(\left(\vec{u} \times \vec{v}\right)\) = \(0\)<br>
\(\Rightarrow\) \(6a^{2} - 42a+36 \) = \(0\)<br>
\(\Rightarrow\) \( a^{2} - 7a +6 \) = \(0\)<br>
\(\Rightarrow\) \(\left(a-6\right)\left(a-1\right)\)<br>
\(\Rightarrow\) \(a\) = \(6\) dan \(a\) = \(1\)<br>
\(a\) = \(6\) \(\Rightarrow\) \(\vec{u} \times \vec{v}\) = \(2\left(6\right)^{3} - 21\left(6\right)^{2} + 36 \left(6\right)\) = \(432-756+216\) = \(-108\)<br>
\(a\) = \(1\) \(\Rightarrow\) \(\vec{u} \times \vec{v}\) = \(2 \left(1\right)^{3} - 21 \left(1\right)^{2} + 36 \left(1\right)\) = \(2-21+36\) = \(17\)<br>
jadi, nilai maksimumnya adalah \(17\)
9. Diketahui vektor - vektor \(\vec{a}\) = \(2\vec{i} + 4\vec{j}+ \vec{k}\), \(\vec{b}\) = \(-3\vec{i} + m\vec{j}+ 2\vec{k}\) dan \(\vec{c}\) = \(\vec{i} + 2\vec{j}- \vec{k}\). Vektor \(\vec{a}\) tegak lurus vektor \(\vec{b}\) maka \(\left(b - c\right)\) adalah
a) \(-4\vec{i} + \vec{j}+3 \vec{k}\)
b) \(-4\vec{i} - \vec{j} + 3\vec{k}\)
c) \(-4\vec{i} -4\vec{j}+ 3\vec{k}\)
d) \(-4\vec{j} + 4\vec{j}+ 3\vec{k}\)
e) \(-4\vec{i} + 3\vec{k}\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}\) = \(2\vec{i} + 4\vec{j}+ \vec{k}\)<br>
\(\vec{b}\) = \(-3\vec{i} + m\vec{j}+ 2\vec{k}\)<br>
\(\vec{c}\) = \(\vec{i} + 2\vec{j}- \vec{k}\)<br>
Vektor \(\vec{a}\) tegak lurus vektor \(\vec{b}\) maka:<br>
\(\vec{a} \times \vec{b}\) = \(0\) \(\Rightarrow\) \(-6+4m+2\)=\(0\)<br>
\(\Rightarrow\) \(m\) = \(1\)<br>
Sehingga, vektor \(\vec{b}\) = \(-3\vec{i} + \vec{j}+ 2\vec{k}\)<br>
jadi, nilai dari:<br>
\(\left(b - c\right)\) = \(\begin {pmatrix} -3 \\ 1 \\ 2 \end {pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\) = \(\begin {pmatrix} -4 \\ -1 \\ 3 \end{pmatrix}\)<br>
=\(-4\vec{i} - \vec{j} + 3\vec{k}\)
10. Agar vektor \(a\) = \(2i + pj + k\) dan \(b = 3i + 2j + 4k\) saling tegakk lurus maka nilai \(p\) adalah
a) \(5\)
b) \(-5\)
c) \(-8\)
d) \(-9\)
e) \(-10\)
Penyelesaian:
Diketahu:<br>
Vektor \(a\) = \(2i + pj + k\) dan \(b = 3i + 2j + 4k\) saling tegak lurus maka:<br>
\(a \times b\) = \(0\)<br>
\(\Rightarrow\) \( 2 \times 3 + p \times 2 + 1 \times 4 \) = \(0\)<br>
\(\Rightarrow\) \(6+2p+4\) = \(0\) \(\Rightarrow\) \(p\) = \(-5\)
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