Question

[7 . 0 . 7]

Bài 1: Cho tứ diện ABCD, biết \(AB^2+CD^2=AD^2+BC^2.\) Hãy tính tích vô hướng \(\overrightarrow{AC.}\overrightarrow{BD}\)

Bài 2: Trong không gian, cho \(\left|\overrightarrow{a}\right|=4;\left|\overrightarrow{b}\right|=5;(\overrightarrow{a};\overrightarrow{b})=120^{\theta}.\) Hãy tính độ dài các vecto sau:

\(a)\left|\overrightarrow{a}-\overrightarrow{b}\right|\)

\(b)\left|2\overrightarrow{a}+\overrightarrow{b}\right|\)

Bài 3: Trong không gian, cho\(\left|\overrightarrow{a}\right|=4;\left|\overrightarrow{b}\right|=3;\overrightarrow{a}.\overrightarrow{b}=10.\) Đặt \(\overrightarrow{x}=\overrightarrow{a}-2\overrightarrow{b};\) \(\overrightarrow{y}=\overrightarrow{a}-\overrightarrow{b}.\) gọi αlà góc giữa hai vecto \((\overrightarrow{x},\overrightarrow{y})\). Hãy tính cosα

Mọi người ơi giúp mình với, mình cảm ơn nhiều ạ!!!

Answer 1

[Nguyễn Việt Lâm]

1/ \(\overrightarrow{AB}^2-\overrightarrow{AD}^2=\overrightarrow{BC}^2-\overrightarrow{CD}^2\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{AB}-\overrightarrow{AD}\right)=\left(\overrightarrow{BC}+\overrightarrow{CD}\right)\left(\overrightarrow{BC}-\overrightarrow{CD}\right)\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{AD}\right).\overrightarrow{DB}=\overrightarrow{BD}\left(\overrightarrow{BC}-\overrightarrow{CD}\right)=\overrightarrow{DB}\left(\overrightarrow{CB}+\overrightarrow{CD}\right)\)

Gọi M là trung điểm BD

\(\Rightarrow2\overrightarrow{AM}.\overrightarrow{DB}=2\overrightarrow{CM}.\overrightarrow{DB}\)

\(\Leftrightarrow\overrightarrow{DB}.\left(\overrightarrow{AM}-\overrightarrow{CM}\right)=0\)

\(\Leftrightarrow\overrightarrow{BD}.\overrightarrow{AC}=0\)

Answer 2

[Nguyễn Việt Lâm]

2/ \(A=\left|\overrightarrow{a}-\overrightarrow{b}\right|\Rightarrow A^2=\overrightarrow{a}^2-2\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}^2\)

\(=a^2+b^2-2ab.cos\left(\overrightarrow{a};\overrightarrow{b}\right)=4^2+5^2-2.4.5.cos120^0=61\)

\(\Rightarrow A=\sqrt{61}\)

b/ \(B=\left|2\overrightarrow{a}+\overrightarrow{b}\right|\Rightarrow B^2=4a^2+b^2+4\overrightarrow{a}.\overrightarrow{b}\)

\(=4a^2+b^2+4ab.cos120^0=49\)

\(\Rightarrow B=7\)

3/ \(\left|\overrightarrow{x}\right|=\left|\overrightarrow{a}-2\overrightarrow{b}\right|\Rightarrow\left|\overrightarrow{x}\right|^2=a^2+4b^2-4\overrightarrow{a}.\overrightarrow{b}=12\)

\(\Rightarrow\left|\overrightarrow{x}\right|=2\sqrt{3}\)

\(\left|\overrightarrow{y}\right|^2=a^2+b^2-2\overrightarrow{a}.\overrightarrow{b}=5\Rightarrow\left|\overrightarrow{y}\right|=\sqrt{5}\)

\(\overrightarrow{x}.\overrightarrow{y}=\left(\overrightarrow{a}-2\overrightarrow{b}\right)\left(\overrightarrow{a}-\overrightarrow{b}\right)=a^2+2b^2-3\overrightarrow{a}.\overrightarrow{b}=4\)

\(\Rightarrow cos\alpha=\frac{\overrightarrow{x}.\overrightarrow{y}}{\left|\overrightarrow{x}\right|.\left|\overrightarrow{y}\right|}=\frac{4}{2\sqrt{15}}=\frac{2\sqrt{15}}{15}\)