Trang chủ
/
Toán
/
Bā 2. Cho |vec(a)|=|vec(b)|=1,(vec(a), vec(b))=45^circ . Túh a, vec(a) cdot vec(b) b,(vec(a)-3 vec(b))(vec(a)+2 vec(b)) c,|vec(a)+vec(b)| d,|vec(a)-sqrt(2) vec(b)|

Câu hỏi

Bā 2. Cho |vec(a)|=|vec(b)|=1,(vec(a), vec(b))=45^circ . Túh a, vec(a) cdot vec(b) b,(vec(a)-3 vec(b))(vec(a)+2 vec(b)) c,|vec(a)+vec(b)| d,|vec(a)-sqrt(2) vec(b)|
zoom-out-in

Bā 2. Cho |vec(a)|=|vec(b)|=1,(vec(a), vec(b))=45^circ . Túh a, vec(a) cdot vec(b) b,(vec(a)-3 vec(b))(vec(a)+2 vec(b)) c,|vec(a)+vec(b)| d,|vec(a)-sqrt(2) vec(b)|

expert verifiedXác minh chuyên gia

Giải pháp

4.5(269 phiếu bầu)
avatar
Chichuyên gia · Hướng dẫn 6 năm

Trả lời

Bài 2<br /><br />Cho $\vert \overrightarrow {a}\vert =\vert \overrightarrow {b}\vert =1$ và $(\overrightarrow {a},\overrightarrow {b})=45^{\circ }$.<br /><br />a) Tính $\overrightarrow {a}\cdot \overrightarrow {b}$<br /><br />$\overrightarrow {a}\cdot \overrightarrow {b} = \vert \overrightarrow {a} \vert \vert \overrightarrow {b} \vert \cos(45^{\circ}) = 1 \cdot 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$<br /><br />b) Tính $(\overrightarrow {a}-3\overrightarrow {b})(\overrightarrow {a}+\overrightarrow {2}\overrightarrow {b})$<br /><br />$(\overrightarrow {a}-3\overrightarrow {b})(\overrightarrow {a}+\overrightarrow {2}\overrightarrow {b}) = \overrightarrow {a}\cdot \overrightarrow {a} + 2\overrightarrow {a}\cdot \overrightarrow {b} - 3\overrightarrow {b}\cdot \overrightarrow {a} - 6\overrightarrow {b}\cdot \overrightarrow {b}$<br /><br />$= \vert \overrightarrow {a} \vert^2 + 2(\overrightarrow {a}\cdot \overrightarrow {b}) - 3(\overrightarrow {a}\cdot \overrightarrow {b}) - 6\vert \overrightarrow {b} \vert^2$<br /><br />$= 1^2 + 2\left(\frac{\sqrt{2}}{2}\right) - 3\left(\frac{\sqrt{2}}{2}\right) - 6(1^2)$<br /><br />$= 1 + \sqrt{2} - \frac{3\sqrt{{2} - 6$<br /><br />$= -5 - \frac{\sqrt{2}}{2}$<br /><br />c) Tính $\vert \overrightarrow {a}-\sqrt {a}\overrightarrow {b}\vert $<br /><br />$\vert \overrightarrow {a}-\sqrt {a}\overrightarrow {b}\vert = \sqrt{(\overrightarrow {a}-\sqrt {a}\overrightarrow {b})\cdot (\overrightarrow {a}-\sqrt {a}\overrightarrow {b})}$<br /><br />$= \sqrt{\vert \overrightarrow {a} \vert^2 - 2\sqrt{a}(\overrightarrow {a}\cdot \overrightarrow {b}) + a\vert \overrightarrow {b} \vert^2}$<br /><br />$= \sqrt{1^2 - 2\sqrt{a}\left(\frac{\sqrt{2}}{2}\right) + a(1^2)}$<br /><br />$= \sqrt{1 - \sqrt{2}a + a}$