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[Wendy Sayuri]

Cho 2 vectơ $\vec{a}, \vec{b}$ thoả mãn: $\left | \vec{a} \right |=1,\left | \vec{b} \right |=2, \left | \vec{a} - 2\vec{b} \right |=\sqrt{15}$  

Xác định k thuộc $\mathbb{R}$ để góc giữa $(\vec{a}+\vec{b}), (2k\vec{a}-\vec{b})$ bằng 60⁰.

[Minhnguyenthe333]

[quote name="Wendy Sayuri" post="574648" timestamp="1437558411"]Cho 2 vectơ $\vec{a}, \vec{b}$ thoả mãn: $\left | \vec{a} \right |=1,\left | \vec{b} \right |=2, \left | \vec{a} - 2\vec{b} \right |=\sqrt{15}$
Xác định k thuộc $\mathbb{R}$ để góc giữa $(\vec{a}+\vec{b}), (2k\vec{a}-\vec{b})$ bằng 60⁰.[/quote
Ta có:$\left | \vec{a} - 2\vec{b} \right |^2=15\Leftrightarrow \vec{a}.\vec{b}=\frac{1}{2}$
$\left | \vec{a} + \vec{b} \right |=\sqrt{6}$
$|2k\vec{a}-\vec{b}|=\sqrt{4k^2-2k+4}$
$( \vec{a} + \vec{b} )(2k \vec{a} -\vec{b})=3k-\frac{9}{2}$
Lại có:$cos60^0=\frac{1}{2}=\frac{( \vec{a} + \vec{b} )(2k \vec{a} -\vec{b})}{| \vec{a} + \vec{b} |.| 2k \vec{a} -\vec{b} |}$
$\Leftrightarrow 6k-9=\sqrt{24k^2-12k+24}$
$\Rightarrow k=...$

Edited by Minhnguyenthe333, 22-07-2015 - 19:16.