Kami kembali memberikan contoh Soal Matematika Vektor Kelas XII SMA Part 4 Semoga bermanfaaat bagi adik adik semua yag masih SMA dan juga para Guru matematika semoga apa yang saya tulis semoga bermanfaat.
a) \(12\vec{i}+12\vec{j}-6\vec{k}\)
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) \(12\vec{i}-12\vec{j}+\vec{k}\)
Penyelesaian:
Diketahui:<br>
\(A\left(2, -1, -3\right)\), \(B\left(-1, 1, -11\right)\) dan \(C\left(4, -3, -2\right)\) maka:<br>
\(\vec{AB}\) = \( \left(-1-2\right)\vec{i} + \left(1-\left(-1\right)\right)\vec{j}+ \left(-11-\left(-3\right)\right)\vec{k}\)<br>
=\(-3 \vec{i}+2\vec{j}-8\vec{k}\)<br>
\(\vec{AC}\) = \( \left(4-2\right)\vec{i} + \left(-3-\left(-1\right)\right)\vec{j}+ \left(-2-\left(-3\right)\right)\vec{k}\)<br>
=\(2 \vec{i}-2\vec{j}+\vec{k}\)<br>
Proyeksi vektor \(\vec{AB}\) pada \(\vec{AC}\)<br>
= \(\frac{\vec{AB} \times \vec{AC}}{\left|\vec{AC}\right|^{2}}\times \vec{AC}\)<br>
=\(\frac{-3(2) +2(-2)+(-8)\times1}{\sqrt{2^{2}+9-2)^{2}+1^{2}}}\times \left(2\vec{i}-2\vec{j}+\vec{k}\right)\)<br>
=\(\frac{-6-4-8}{\left(\sqrt{4+4+1}\right)^{2}}\times \left(2\vec{i}-2\vec{j}+\vec{k}\right)\)<br>
=\(\frac{-18}{9}\times \left(2\vec{i}-2\vec{j}+\vec{k}\right)\) = \(-4\vec{i}+4\vec{j}-2\vec{k}\)
42. Diketahui titik - titik \(P\left(1,1\right)\), \(Q\left(5,3\right)\), dan \(R\left(2,4\right)\). Jika titik \(5\) merupakan proyeksi titik \(R\) pada garis \(PQ\). Maka panjang \(PS\)=...
a) \(\frac{\sqrt{5}}{5}\)
b) \(\frac{\sqrt{5}}{3}\)
c) \(\frac{2\sqrt{5}}{5}\)
d) \(\frac{\sqrt{5}}{2}\)
e) \(\sqrt{5}\)
Penyelesaian:
\(\vec{PQ}\) = \(\left(5,3\right) - \left(1,1\right)\) = \(\left(4, 2\right)\)<br>
\(\vec{PR}\) = \(\left(2,4\right) - \left(1,1\right)\) = \(\left(1,3 \right)\)<br>
\(\left|\vec{PS}\right|\) = \(\frac{\vec{PQ}\times \vec{PR}}{\left|PQ\right|}\) = \(\frac{1 \times 4 + 2 \times 3}{\sqrt{4^{2}+2^{2}}}\)<br>
=\(\frac{4+6}{\sqrt{16+4}}\) = \(\frac{10}{\sqrt{20}}\) = \(\frac{10}{2\sqrt{5}}\)<br>
=\(\frac{5}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}\) = \(\sqrt{5}\)
43. Diketahui \(\vec{a}+\vec{b}\) = \(\vec{i}-\vec{j}+4\vec{k}\) dan \(\left|\vec{a}-\vec{b}\right|\) = \(\sqrt{14}\). Hasil dari \(\vec{a} \times \vec{b}\)=...
a) \(4\)
b) \(2\)
c) \(1\)
d) \(\frac{1}{2}\)
e) \(0\)
Penyelesaian:
Diketahui:<br>
\(\vec{a}+\vec{b}\) = \(\vec{i}-\vec{j}+4\vec{k}\)<br>
\(\left|\vec{a}+\vec{b}\right|\) = \(\sqrt{1^{2}+\left(-1\right)^{2}+4^{2}}\) = \(\sqrt{18}\)<br>
\(\Rightarrow\) \(\left|\vec{a}+\vec{b}\right|^{2}\) = \(18\)<br>
\(\Rightarrow\) \(\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}+2\vec{a} \times \vec{b}\) = \(18\)... \(\left(i\right)\)<br>
\(\Rightarrow\) \(\left|\vec{a}-\vec{b}\right|\) = \(\sqrt{14}\)<br>
\(\Rightarrow\) \(\left|\vec{a}-\vec{b}\right|^{2}\) = \(14\)<br>
\(\Rightarrow\) \(\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}-2\vec{a} \times \vec{b}\) = \(14\)... \(\left(ii\right)\)<br>
Dari \(\left(i\right)\) dan \(\left(ii\right)\):<br>
\(\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}+2\vec{a} \times \vec{b}\) = \(18\)<br>
\(\left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}-2\vec{a} \times \vec{b}\) = \(14\)<br>
Jika dikurangi \(4\vec{a} \times \vec{b}\) = \(4\)<br>
\(\Rightarrow\) \(\vec{a} \times \vec{b}\) = \(1\)
44. Sudut antar vektor <br>
\(\vec{a}\) = \(2x+1 \vec{i}-\vec{j}+4\vec{k}\)= \(\angle\) \(60^{\circ}\) . Jika panjang proyeksi \(\vec{a}\) ke \(\vec{b}\) sama dengan \(\frac{1}{2}\)
a) \(4\) atau \(\frac{1}{2}\)
b) \(1\) atau \(4\)
c) \(\frac{6}{6}\sqrt{7}\)
d) \(\sqrt{6}\)
e) \(-\frac{1}{2}\) atau \(1\)
Penyelesaian:
Vektor \(\vec{p}\) dan vektor \(\vec{q}\) membentuk sudut \(\alpha\), berlaku \(\alpha\) \(\cos\) \(\alpha\) = \(\frac{\vec{p} \times \vec{q}}{\left|\vec{p}\right|\left|\vec{p}\right|}\)<br>
Panjang vektor:<br>
\(\vec{p}\) = \(\left|\vec{p}\right|\) = \(\sqrt{\left(-3\right)^{2} + 3^{2} + 0^{2}}\) = \(3\sqrt{2}\)<br>
Panjang vektor:<br>
\(\vec{q}\) = \(\left|\vec{q}\right|\) = \(\sqrt{2^{2} + 3^{2} + \left(-2\right)^{2}}\) = \(\sqrt{14}\)<br>
\(\cos\alpha\) = \(\frac{\begin{pmatrix} -3 \\ 3 \\ 0 \end {pmatrix}\vec{q} = \begin{pmatrix} 1 \\ 3 \\ -2 \end {pmatrix}}{3\sqrt{2} \times \sqrt{14}}\) = \(\frac{-3+9+0}{6\sqrt{7}}\) = \(\frac{1}{\sqrt{7}}\)<br>
Dengan phtagoras:<br>
\(\sqrt{\left(\sqrt{7}\right)^{2}}\) = \(\sqrt{6}\)<br>
Jadi, \(\tan \alpha\)=\(\sqrt{6}\)
45. Diketahui \(\vec{p}\) = \(\begin{pmatrix} -3 \\ 3 \\ 0 \end {pmatrix}\) dan \(\vec{q}\) = \(\begin{pmatrix} 1 \\ 3 \\ -2 \end {pmatrix}\)<br>
apabila \(\alpha\) adalah sudut yang dibentuk antara vektor \(\vec{p}\) dan \(\vec{q}\), maka \(\tan\alpha\)=...
a) \(\frac{1}{6}\sqrt{6}\)
b) \(\frac{1}{7}\sqrt{7}\)
c) \(\frac{6}{6}\sqrt{7}\)
d) \(\sqrt{6}\)
e) \(\sqrt{6}\)
Penyelesaian:
\(\vec{p}\) = \(\begin{pmatrix} -3 \\ 3 \\ 0 \end {pmatrix}\) dan \(\vec{q}\) = \(\begin{pmatrix} 1 \\ 3 \\ -2 \end {pmatrix}\)
\(\tan\alpha\)=\(\sqrt{6}\)
46. 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>
\(\vec{s}\) = \(-\vec{p}+\vec{q}+\vec{r}\)
47. \(ABCD\) adalah belah ketupat. Jika \(\vec{AD}\) = \(\vec{u}\), \(\vec{AB}\) = \(\vec{v}\) dan besar sudut \(BAD\) adalah \(\alpha\) maka akan selalu=..
a) \(\vec{u}\) tegak lurus pada \(\vec{v}\)
b) \(\left|\vec{u}+\vec{v}\right|\) = \(2 \left|\vec{u}\right|\) atau \(\left|\vec{u}+\vec{v}\right| \)= \(2 \left|\vec{v}\right|\)
c) Proyeksi \(\vec{u}\) pada \(\vec{v}\) adalah \(\vec{u} \sin \alpha\)
d) \(\vec{u}+\vec{u}\) tegak lurus \(\vec{u}-\vec{v}\)
e) \(\left|\vec{u}+\vec{v}\right|\) = \(2\left|\vec{u}\right|\) atau \(\left|\vec{u}+\vec{v}\right| \)= \(2\left|\vec{v}\right|\)
Penyelesaian:
Pernyataan \(1)\) salah, karena \(\vec{u}\) tidak tegak lurus pada \(\vec{v}\)<br>
Pernyataan \(2)\) salah, karena:<br>
\(\left|\vec{u} + \vec{v}\right|\) =\(\sqrt{\left|\vec{v}\right|^{2}+\left|\vec{u}\right|^{2} + 2\left|\vec{u}\right|\left|\vec{v}\right|}\) sehingga<br>
\(\left|\vec{u}+\vec{v}\right|\) tidak sama dengan \(2 \left|\vec{u}\right|\) tidak sama dengan \(2 \left|\vec{v}\right|\)<br>
Pernyataan \(3)\) salah, karena proyeksio\ \(\vec{u}\) pada \(\vec{f}\) harus searah dengan \(\vec{v}\).<br>
Pernyataan \(4)\) benar karena :<br>
\(\vec{u}+\vec{v}\)\vec{u}-\vec{v}))\) = \(\left|\vec{u}\right|^{2} - \left|\vec{v}\right|^{2}\) = karena \(\left|\vec{u}\right|\) = \(\left|\vec{v}\right|\)
48. Pada segiempat sembarang \(ABCD\). \(S\) dan \(T\) masong - masing titik tengah \(AC\) dan \(BD\). Jika \(\vec{u}\) = \(\vec{ST}\) maka \(\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}\) dapat dinyatakan sama \(\vec{u}\) sebagai...
a) \(\frac{1}{4}\vec{u}\)
b) \(\frac{1}{2}\vec{u}\)
c) \(\vec{u}\)
d) \(2\vec{u}\)
e) \(4\vec{u}\)
Penyelesaian:
\(T\) tengah - tengah \(BD\) maka:<br>
\(\vec{AT}\) = \(\frac{1}{2}\left(\vec{AB}+\vec{AD}\right)\) .... \(\left(i\right)\)<br>
\(\vec{CT}\) = \(\frac{1}{2}\left(\vec{CB}+\vec{CD}\right)\) .... \(\left(ii\right)\)<br>
\(\vec{AT}+\vec{CT}\) =\(\left(\vec{AS}+\vec{ST}\right) + \left(\vec{CS}+\vec{ST}\right)\)<br>
=\(\vec{AS}+\vec{CS}+2\vec{ST}\)<br>
=\(0+\vec{CS}+2\vec{ST}\)<br>
Jadi, \(\vec{ST}\) maka \(\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}\) = \(2\left(\vec{AT}+\vec{CT}\right)\)<br>
=\(4\vec{ST}\) = \(4\vec{u}\)<br>
49. Diketahui vektor \(\vec{a}\) = \(2i – 2pj +4k\) dan \(\vec{b}\) = \(i – 3j +4k\). Jika panjang proyeksi vektor \(\vec{a}\) pada \(\vec{b}\) adalah \(\frac{6}{\sqrt{26}}\). Nilai \(p\) = ....
a) \(-3\)
b) \(-2\)
c) \(-1\)
d) \(1\)
e) \(3\)
Penyelesaian:
\(\left|\vec{b}\right|\) = \(\sqrt{1^{2}+ \left(-3\right)^{2} + 4^{2}}\) = \(\sqrt{26}\)<br>
Panjang proyeksi vektor \(\vec{a}\) pada \(\vec{b}\) = \(\frac{\vec{a} \times \vec{b}}{\left|\vec{b}\right|}\)<br>
\(\frac{6}{\sqrt{26}}\)= \(\frac{\begin{pmatrix}2\\-2p\\4\end{pmatrix}\begin{pmatrix}1\\-3\\4\end{pmatrix}}{26}\)<br>
\(6\) = \(2 + 6p + 16\) \(\rightarrow\) \(p\) = \(-2\)
50. Diketahui segitiga \(ABC\) dengan \(A\left(2, 1, 2\right)\), \(B\left(6, 1, 2\right)\), dan \(C\left(6, 5, 2\right)\). Jika \(\vec{u}\) mewakili \(\vec{AB}\) dan \(\vec{v}\) mewakili \(\vec{AC}\), maka sudut yang dibentuk oleh vektor \(\vec{u}\) dan \(\vec{v}\) adalah ...
a) \(30^{\circ}\)
b) \(\sqrt{19}\)
c) \(4\sqrt{7}\)
d) \(2\sqrt{7}\)
e) \(120^{\circ}\)
Penyelesaian:
Diketahui \(\left|\vec{a}\right|\)=\(4\) dan \(\left|\vec{b}\right|\) = \(6\)<br>
\(\left|\vec{a} - \vec{b} \right|\) = \(2\sqrt{19}\)<br>
\(\sqrt{\left(\left|\vec{a} - \vec{b}\right|\right)^{2}}\) = \(2\sqrt{19}\)<br>
\(\vec{a} \times \vec{a} + \vec{b} \times \vec{b} - 2\vec{a} \times \vec{b} \) = 76<br>
\(|\vec{a}|^{2} +|\vec{b}|^{2} - 2 (|\vec{a}|) (|\vec{b}|) \cos x \) = \(76\)<br>
\(16 +36 - 2 \left(4\right) \left(6\right) \cos x\) = \(76\)<br>
\(\cos x\) = \(\frac{76-16-36}{-48}\) = \(\frac-{24}{48}\) = \(\frac-{1}{2}\)<br>
\(\left|\vec{a} + \vec{b} \right|\) = \(\sqrt{\left(\left|\vec{a} + \vec{b}\right|\right)^{2}}\)<br>
=\(\sqrt{\vec{a} \times \vec{a} + \vec{b} \times \vec{b} + 2\vec{a} \times \vec{b}} \) <br>
=\(\sqrt{\left|\vec{a}\right|^{2} +\left|\vec{b}\right|^{2} + 2 \left(\left|\vec{a}\right|\right) (|\vec{b}|) \cos x} \) <br>
=\(\sqrt{16+36+2 \times 4 \times 6(-\frac{1}{2}}\)<br>
=\(\sqrt{28}\)<br>
=\(2\sqrt{7}\)
Terima kasih telah mengunjungi website kami. Next contoh soal dan pembahasan selanjutnya .........