реши 7 8 и 9
Evaluating geometry problems
I need to provide solutions in Russian, starting with brief answers for the problems related to the rhombus and the triangle. For the rhombus, I’ll determine the lengths of diagonals based on their intersection point and calculate the dot product of vectors AO and BO. Since the diagonals are perpendicular, their dot product will equal zero. For the triangle, I’ll find the length of the resultant vector using the cosine rule. The final answer is 6.Summarizing vector calculations
I need to note that the sum of vectors \(\vec{AB} + \vec{AC}\) leads to the midpoint and that its length is 6, which is correct. For the rectangle problem, the adjacent sides AB and AD are perpendicular, so their dot product is 0, and I should clarify that vectors from point A confirm the right angle. I’ll structure the response to include concise answers with brief explanations. The final note could be: "Разберём похожее задание на сумму векторов?"
Краткие ответы:
7) \(0\); 8) \(6\); 9) \(0\).
7
Диагонали ромба перпендикулярны и делятся точкой пересечения пополам. Векторы \(\vec{AO}\) и \(\vec{BO}\) лежат на разных диагоналях, поэтому угол между ними равен \(90^\circ\).
\[ \vec{AO}\cdot\vec{BO} =AO\cdot BO\cdot\cos 90^\circ=0. \]
Ответ: \(0\).
8
В равностороннем треугольнике:
\[ AB=AC=2\sqrt3,\qquad \angle BAC=60^\circ. \]
По формуле квадрата длины суммы векторов:
\[ |\vec{AB}+\vec{AC}|^2 =AB^2+AC^2+2\cdot AB\cdot AC\cos60^\circ. \]
\[ |\vec{AB}+\vec{AC}|^2 =(2\sqrt3)^2+(2\sqrt3)^2+ 2(2\sqrt3)(2\sqrt3)\cdot\frac12 =12+12+12=36. \]
\[ |\vec{AB}+\vec{AC}|=\sqrt{36}=6. \]
Ответ: \(6\).
9
Стороны прямоугольника \(AB\) и \(AD\) перпендикулярны:
\[ \angle BAD=90^\circ. \]
По формуле скалярного произведения:
\[ \vec{AB}\cdot\vec{AD} =AB\cdot AD\cdot\cos90^\circ=0. \]
Ответ: \(0\).
