نمونه سوالات فصل 5 ریاضی هشتم همراه با جواب تشریحی

از تعریف بردارها استفاده می کنیم:

\(\begin{array}{l}1)\,\overrightarrow {AB} = B - \,A = \left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right]\, - \left[ {\begin{array}{*{20}{c}}4\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2 - 4}\\{3 - 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 6}\\0\end{array}} \right]\\\\\\2)\,\overrightarrow {AB} = B - \,A = \left[ {\begin{array}{*{20}{c}}x\\3\end{array}} \right]\, - \left[ {\begin{array}{*{20}{c}}2\\y\end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{x - 2}\\{3 - y}\end{array}} \right]\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x - 2 = 5}\\{3 - y = 1}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x = 7}\\{y = 2}\end{array}} \right. \Rightarrow A = \left[ {\begin{array}{*{20}{c}}2\\2\end{array}} \right]\,\,,B = \left[ {\begin{array}{*{20}{c}}7\\3\end{array}} \right]\end{array}\)

1) بردار \(\left[ {\begin{array}{*{20}{c}}5\\0\end{array}} \right]\)، عرضش صفر است؛ پس موازی محور طول ها است.

2) بردار \(\left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\) هیچ یک از مؤلفه هایش صفر نیست؛ پس موازی هیچ کدام از محورها نیست.

3) برای اینکه برداری موازی محور y باشد، باید طول آن صفر باشد؛ پس \(\left[ {\begin{array}{*{20}{c}}0\\{ - 4}\end{array}} \right]\) موازی محور y است.

ابتدا بردار \(\vec a = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\) با نقطه ابتدا مبدأ را رسم می کنیم که یعنی از مبدأ دو واحد به راست و یک واحد به پایین می رویم:

برای رسم \(3\vec a\) و \( - \vec a\) می توانیم مختصات \(\vec a\) را در 3 و \( - 1\) ضرب کنیم؛ یعنی \(\;3\vec a = \left[ {\begin{array}{*{20}{c}}{3 \times 2}\\{3 \times \left( { - 1} \right)}\end{array}} \right]\) و \( - \vec a = \left[ {\begin{array}{*{20}{c}}{\left( { - 1} \right) \times 2}\\{\left( { - 1} \right) \times \left( { - 1} \right)}\end{array}} \right]\)؛ در نتیجه:

\(\;3\vec a = \left[ {\begin{array}{*{20}{c}}6\\{ - 3}\end{array}} \right]\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\; - \vec a = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right]\)

این دو بردار را جداگانه رسم می کنیم:

\(\begin{array}{l}1)\,\,\frac{1}{3}\left[ {\begin{array}{*{20}{c}}9\\{ - 6}\end{array}} \right] = \,\left[ {\begin{array}{*{20}{c}}{\frac{9}{3}}\\{\frac{{ - 6}}{3}}\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}3\\{ - 2}\end{array}} \right] = 3\vec i - 2\vec j\\\\2)\,\,2\vec i - 3\vec j + \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right] = 2\vec i - 3\vec j + 5\vec i + 4\vec j = \\\left( {2 + 5} \right)\vec i + \left( { - 3 + 4} \right)\vec j = 7\vec i + \vec j\\\\3)\,\,\left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right] - 2\vec j = 2\vec i + 0\vec j - 2\vec j = 2\vec i - 2\vec j\\\\4)\,\,\left[ {\begin{array}{*{20}{c}}0\\3\end{array}} \right] - 3\vec j = 0\vec i + 3\vec j - 3\vec j = 0\vec i + 0\vec j\end{array}\)

دو بردار مساوی دارای طول و عرض مساوی هستند.

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{5m - 2}\\{3n + 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8\\{2n}\end{array}} \right] \Rightarrow \left\{ {\begin{array}{*{20}{c}}{5m - 2 = 8}\\{3n + 1 = 2n}\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}}{5m = 10}\\{3n - 2n = - 1}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{m = 2}\\{n = - 1}\end{array}} \right.\end{array}\)

\(\begin{array}{l}1)\,\,2\vec x + \vec i = \left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right] \Rightarrow 2\vec x = \left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right] - \vec i = \\\left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\6\end{array}} \right] \Rightarrow \vec x\frac{1}{2}\left[ {\begin{array}{*{20}{c}}4\\6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\\\\2)\,\,\vec i + \vec x = \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right] \Rightarrow \vec x = \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right] - \vec i = \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ \Rightarrow \vec x = \left[ {\begin{array}{*{20}{c}}2\\5\end{array}} \right]\\\\3)\,\,3\vec i + 2\vec j + 2\vec x = \left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right] \Rightarrow 2\vec x = \left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right] - 3\vec i - 2\vec j = \\\left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\4\end{array}} \right] \Rightarrow \vec x = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}2\\4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\end{array}\)

\(1)\,\vec b + \vec c = \vec a\)

\(2)\,\vec a + \vec b + \vec c = \vec d\)

توضیح 1) از A دو خط مشخص شده رسم می کنیم تا نقاط B و C مشخص شوند؛ پس بردارهای OB و OC تجزیه بردار OA هستند.

توضیح 2) \(\overrightarrow {OC} + \overrightarrow {OB} = \overrightarrow {OA} \)

\(1)\,\vec c = \vec a + \vec b\)

\(2)\,\vec d = - 2\vec a + 3\vec b \)

بردار  را 3 بار به نقطه A اضافه می کنیم تا نقطه B به دست آید:

\(\begin{array}{l}B = A + 3\vec v = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}3\\{ - 6}\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{1 + 3}\\{3 - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\{ - 3}\end{array}} \right]\end{array}\)

مجموع بردارهای \({\overrightarrow F _1}\) و \({\overrightarrow F _2}\) را به دست می آوریم (به روش جمع برداری) تا جهت حرکت یعنی همان \({\overrightarrow F _1} + {\overrightarrow F _2}\) به دست آید:

از رابطه بردارها استفاده می کنیم:

نقطه ابتدا – نقطه انتها = بردار

\(\vec c = B - A = \left[ {\begin{array}{*{20}{c}}3\\4\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 1}\\{4 - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\2\end{array}} \right]\)

\(\begin{array}{l}B = A + \vec a = \left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 + 1}\\{ - 4 + 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right]\\\\C = B + \vec b = \left[ {\begin{array}{*{20}{c}}4\\{ - 2}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4 - 2}\\{ - 2 + 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\end{array}\)

\(\begin{array}{l}\vec c = 2\vec a - 3\vec b = 2\left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}1\\{ - 4}\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{2 \times 2}\\{3 \times 2}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{1 \times 3}\\{ - 4 \times 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4\\6\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3\\{ - 12}\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{4 - 3}\\{6 + 12}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{18}\end{array}} \right]\end{array}\)

بردار \(\left[ {\begin{array}{*{20}{c}}4\\{ - 1}\end{array}} \right]\) بعنی چهار واحد به سمت راست و یک واحد به سمت پایین؛ یعنی به صورت:

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right] + 3\vec x = 3\vec i + 5\vec j - \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] - \vec x\\\\ \Rightarrow 3\vec x + \vec x = 3\vec i + 5\vec j - \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right]\\\\ \Rightarrow 4\vec x = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{3 + 0 - 2 - 2}\\{0 + 5 - 1 - \left( { - 4} \right)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right]\\\\ \Rightarrow \vec x = \frac{1}{4}\left[ {\begin{array}{*{20}{c}}{ - 1}\\8\end{array}} \right] \Rightarrow \vec x = \left[ {\begin{array}{*{20}{c}}{ - \frac{1}{4}}\\2\end{array}} \right]\end{array}\)

\(\begin{array}{l}1)\left[ {\begin{array}{*{20}{c}}{ - 5}\\6\end{array}} \right] = - 5\vec i + 6\vec j\\\\2)\left[ {\begin{array}{*{20}{c}}3\\{ - 4}\end{array}} \right] = 3\vec i - 4\vec j\end{array}\)

\(\begin{array}{l}1)\,3\vec i - \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] + 4\vec j = \left[ {\begin{array}{*{20}{c}}3\\0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\4\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{3 - 2 + 0}\\{0 - 1 + 4}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]\\\\2)\, - 5\vec j + 3\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right] - 2\vec i = \left[ {\begin{array}{*{20}{c}}0\\{ - 5}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{1 \times 3}\\{2 \times 3}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{0 + 3 - 2}\\{ - 5 + 6 - 0}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\end{array}\)

مساوی

راستا، جهت، اندازه

موازی

برابر

صفر

صفر

نیمساز ناحیه اول و سوم

نیمساز ناحیه دوم و چهارم

\(\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\)

جمع برداری

مثلث، متوازی الاضلاع

\(\left[ {\begin{array}{*{20}{c}}{kx}\\{ky}\end{array}} \right]\)

تجزیه بردار

بردار

\(\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right]\)

طول ها (یا x ها)

عرض ها (یا y ها)

\(\left[ {\begin{array}{*{20}{c}}{ - \frac{3}{2}}\\{\frac{5}{2}}\end{array}} \right]\)

راست، پایین

هر بردار دارای راستا، جهت و اندازه است.

دو بردار \(\vec a\) و \( - \vec a\) که با هم مختلف الجهت هستند، با هم موازی هستند؛ جهت \( - \vec a\) دقیقاً برعکس جهت \(\vec a\) می باشد.

موازی نیم ساز ناحیه دوم و چهارم است.

بردار \(\vec j\)، به صورت \(\left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\) می باشد.

موازی محور طول ها است.

\(\left[ {\begin{array}{*{20}{c}}4\\3\end{array}} \right] - 5\vec i + 3\vec j = \left[ {\begin{array}{*{20}{c}}4\\3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 5}\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}\\6\end{array}} \right]\)

\(\begin{array}{l}\overrightarrow {AB} = B - A = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 3}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 4 - 3}\\{ - 3 - 2}\end{array}} \right] = \\\left[ {\begin{array}{*{20}{c}}{ - 7}\\{ - 5}\end{array}} \right]\end{array}\)

\(\begin{array}{l}\vec a = \vec b \Rightarrow \left[ {\begin{array}{*{20}{c}}{x - 1}\\{y + 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3x - 5}\\{2y - 5}\end{array}} \right]\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x - 1 = 3x - 5}\\{y + 3 = 2y - 5}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2x = 4}\\{y = 8}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x = 2}\\{y = 8}\end{array}} \right.\\\\ \Rightarrow \vec a\left[ {\begin{array}{*{20}{c}}{2 - 1}\\{8 + 3}\end{array}} \right] \Rightarrow \vec a\left[ {\begin{array}{*{20}{c}}1\\{11}\end{array}} \right]\end{array}\)

اگر برداری موازی محور عرض ها باشد، یعنی طول آن بردار صفر است:

\(3a + 6 = 0\,\,\,\, \Rightarrow \,\,\,\,3a = - 6\,\,\,\, \Rightarrow \,\,\,\,a = - 2\)

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{2m - 1}\\3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3m + 5}\\3\end{array}} \right]\,\,\, \Rightarrow \,\,2m - 1 = - 3m + 5\\\\ \Rightarrow \,\,5m = 5 + 1 = 6\,\,\, \Rightarrow \,\,\,m = \frac{6}{5}\end{array}\)

وقتی یک بردار موازی محور طول ها باشد، عرض آن صفر است:

\(1 - 2n = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,2n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,n = \frac{1}{2}\)

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 7}\\8\end{array}} \right]\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x + 2 = - 7}\\{y - 3 = 8}\end{array}} \right.\\\\ \Rightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = - 7 - 2}\\{y = 8 + 3}\end{array}} \right.\,\,\, \Rightarrow \,\,\,\left\{ {\begin{array}{*{20}{c}}{x = - 9}\\{y = 11}\end{array}} \right.\end{array}\)

ابتدا بردار \(\vec c\) را با استفاده از دو بردار \(\vec a\) و \(\vec b\) محاسبه می کنیم:

\(\vec c = 3\vec a - 2\vec b = 3\left[ {\begin{array}{*{20}{c}}4\\1\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{12 - 4}\\{3 + 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right]\)

بردار \(\vec c\)، یک بردار با طول و عرض مثبت است که اندازه طول آن بیشتر از عرض آن است؛ پس شکل تقریبی آن به صورت زیر خواهد بود.

\(\left[ {\begin{array}{*{20}{c}}0\\{ - 6}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right]\)

باید بردار انتقال \(\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} \) را پیدا کنیم:

\(\begin{array}{l}A + \overrightarrow {AC} = C\,\,\, \Rightarrow \,\,\,C - A = \overrightarrow {AC} \\\\ \Rightarrow \,\,\,\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} = \left[ {\begin{array}{*{20}{c}}{ - 5}\\7\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6\\{ - 8}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]\end{array}\)

\(\begin{array}{l}\frac{2}{3}\left[ {\begin{array}{*{20}{c}}9\\{15}\end{array}} \right] - \frac{1}{5}\left[ {\begin{array}{*{20}{c}}5\\{20}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\frac{2}{3} \times 9}\\{\frac{2}{3} \times 15}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{\frac{1}{5} \times 5}\\{\frac{1}{5} \times 20}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}6\\{10}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1\\4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\6\end{array}} \right]\end{array}\)

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right] + 15\left[ {\begin{array}{*{20}{c}}{0/4}\\{ - 0/2}\end{array}} \right] + 10\left[ {\begin{array}{*{20}{c}}{ - 0/3}\\{0/5}\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6\\{ - 3}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 3}\\5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5\\2\end{array}} \right]\end{array}\)

\( - \left[ {\begin{array}{*{20}{c}}3\\5\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\end{array}} \right]\)

اگر برداری بر محور طول ها عمود باشد، طول بردار برابر صفر است؛ بنابراین:

\(\begin{array}{l}a + 1 = 0\,\, \Rightarrow \,a = - 1\,\, \Rightarrow \,\vec m = \left[ {\begin{array}{*{20}{c}}{a - 1}\\{1 + a}\end{array}} \right] = \\\\\left[ {\begin{array}{*{20}{c}}{ - 1 - 1}\\{1 + \left( { - 1} \right)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right] = - 2\vec i\end{array}\)

با توجه به شکل، واضح است که مختصات بردارهای \(\vec a\) به \(\vec b\) صورت زیر است:

\(\begin{array}{l}\left. {\begin{array}{*{20}{c}}{\vec a = 3\vec i + 3\vec j}\\{\vec b = - \vec i + 2\vec j}\end{array}} \right\} \Rightarrow \,\,\vec c = 2\vec i + \vec a - \vec b = \\\\2\vec i + 3\vec i + 3\vec j - \left( { - \vec i + 2\vec j} \right) = \\\\5\vec i + 3\vec j + \vec i - 2\vec j = 6\vec i + \vec j\end{array}\)

\(2\vec i + \left[ {\begin{array}{*{20}{c}}0\\5\end{array}} \right] - 3\vec j = 2\vec i + 5\vec j - 3\vec j = 2\vec i + 2\vec j\)

\( - \left[ {\begin{array}{*{20}{c}}5\\{ - 4}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{ + 4}\end{array}} \right] = - 5\vec i + 4\vec j\)

\(2\vec b - 6\vec a = 2\left( {3\vec a} \right) - 6\vec a = 6\vec a - 6\vec a = \vec o\)

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\end{array}} \right] + 3\vec x = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\,\,\, \Rightarrow \,\,3\vec x = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{5 + 1}\\{4 + 5}\end{array}} \right]\\\\ \Rightarrow \,3\vec x = \left[ {\begin{array}{*{20}{c}}6\\9\end{array}} \right]\,\,\, \Rightarrow \,\,\vec x = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}6\\9\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\end{array}\)

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{1 - m - 3 = 5}\\{n + 2 - \left( { - 2} \right) = 1}\end{array}} \right.\,\, \Rightarrow \left\{ {\begin{array}{*{20}{c}}{ - m - 2 = 5}\\{n + 4 = 1}\end{array}} \right.\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{c}}{ - m = 7}\\{n = - 3}\end{array}} \right.\,\, \Rightarrow \left\{ {\begin{array}{*{20}{c}}{m = - 7}\\{n = - 3}\end{array}} \right. \Rightarrow m + n = - 10\end{array}\)