Bài tập Tính độ dài đoạn thẳng, độ dài vectơ lớp 10 (có lời giải)

Hướng dẫn giải:

Đáp án đúng là: B.

$\overrightarrow{a}.\overrightarrow{b}=2\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a},\overrightarrow{b}\right)=2$

$\iff 2.\left|\overrightarrow{b}\right|.cos60°=2\iff 2.\left|\overrightarrow{b}\right|.\frac{1}{2}=2\iff \left|\overrightarrow{b}\right|=2$.

Hướng dẫn giải:

Đáp án đúng là: D.

$\overrightarrow{a}.\overrightarrow{b}=4\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos\left(\overrightarrow{a},\overrightarrow{b}\right)=4\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.cos60°=4$

$\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.\frac{1}{2}=4\iff \left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|=8$.

Ta có: $\left\{\begin{array}{l}\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|=8\\ \left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|=6\end{array}\right.\iff \left\{\begin{array}{l}\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|=8\\ \left|\overrightarrow{b}\right|=6-\left|\overrightarrow{a}\right|\end{array}\right.\iff \left\{\begin{array}{l}\left|\overrightarrow{a}\right|.\left(6-\left|\overrightarrow{a}\right|\right)=8\left(1\right)\\ \left|\overrightarrow{b}\right|=6-\left|\overrightarrow{a}\right|\end{array}\right.$

Xét (1) ta có : $6\left|\overrightarrow{a}\right|-{\left|\overrightarrow{a}\right|}^2=8\iff -{\left|\overrightarrow{a}\right|}^2+6\left|\overrightarrow{a}\right|-8=0\iff \left[\begin{array}{l}\left|\overrightarrow{a}\right|=4\\ \left|\overrightarrow{a}\right|=2\end{array}\right.$

Vì $\left|\overrightarrow{a}\right|>3$ nên ta có: $\left|\overrightarrow{a}\right|=4$.

Hướng dẫn giải:

Đáp án đúng là: B.

Ta có:

$B{C}^2={\overrightarrow{BC}}^2={\left(\overrightarrow{AC}-\overrightarrow{AB}\right)}^2={\overrightarrow{AC}}^2-2\overrightarrow{AC}.\overrightarrow{AB}+{\overrightarrow{AB}}^2$

$=A{C}^2-2\overrightarrow{AC}.\overrightarrow{AB}+A{B}^2$.

Ta có:

$\cos \left(\overrightarrow{AC},\overrightarrow{AB}\right)=\cos \widehat{A}=cos45°=\frac{\sqrt{2}}{2}$

$\left|\overrightarrow{AC}\right|=AC=3;\left|\overrightarrow{AB}\right|=AB=2$

$\overrightarrow{AC}.\overrightarrow{AB}=\left|\overrightarrow{AC}\right|.\left|\overrightarrow{AB}\right|.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=\mathrm{3.2.}\frac{\sqrt{2}}{2}=3\sqrt{2}$

Do đó, ta có:

$B{C}^2=A{C}^2-2\overrightarrow{AC}.\overrightarrow{AB}+A{B}^2=3^2-2.3\sqrt{2}+2^2=13-6\sqrt{2}$

$\Rightarrow BC=\sqrt{13-6\sqrt{2}}$.

Hướng dẫn giải:

Đáp án đúng là: D.

Ta có:

$C{M}^2={\overrightarrow{CM}}^2={\left(\overrightarrow{BC}-\overrightarrow{BM}\right)}^2={\overrightarrow{BC}}^2-2\overrightarrow{BC}.\overrightarrow{BM}+{\overrightarrow{BM}}^2$

$=B{C}^2-2\overrightarrow{BC}.\overrightarrow{BM}+B{M}^2$

Ta có:

$cos\left(\overrightarrow{BC},\overrightarrow{BM}\right)=\cos \widehat{CBM}=cos60°=\frac{1}{2}$.

$\left|\overrightarrow{BC}\right|=BC=2cm$; $\left|\overrightarrow{BM}\right|=BM=1cm$

Ta có: $\overrightarrow{BC}.\overrightarrow{BM}=\left|\overrightarrow{BC}\right|.\left|\overrightarrow{BM}\right|.cos\left(\overrightarrow{BC},\overrightarrow{BM}\right)=\mathrm{2.1.}\frac{1}{2}=1$

$C{M}^2=B{C}^2-2\overrightarrow{BC}.\overrightarrow{BM}+B{M}^2=2^2-2.1+1^2=3\Rightarrow CM=\sqrt{3}$ (cm)

Mà M là trung điểm của AC

$\Rightarrow AC=2CM=2\sqrt{3}$ cm.

Hướng dẫn giải:

Đáp án đúng là: A.

AM là trung tuyến nên M là trung điểm của BC.

Do đó, có: $\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AM}\Rightarrow 4{\overrightarrow{AM}}^2={\left(\overrightarrow{AB}+\overrightarrow{AC}\right)}^2$

$\iff 4A{M}^2=A{B}^2+2\overrightarrow{AB}.\overrightarrow{AC}+A{C}^2$.

$\cos \left(\overrightarrow{AB},\overrightarrow{AC}\right)=\cos \widehat{A}=\cos 120°=-\frac{1}{2}$

$\left|\overrightarrow{AB}\right|=AB=4$

$\left|\overrightarrow{AC}\right|=AC=3$

$\overrightarrow{AB}.\overrightarrow{AC}=\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|.\cos \left(\overrightarrow{AB},\overrightarrow{AC}\right)=\mathrm{4.3.}\left(\frac{-1}{2}\right)=-6$

$4A{M}^2=A{B}^2+2\overrightarrow{AB}.\overrightarrow{AC}+A{C}^2=4^2+2.(-6)+3^2=13$

$\Rightarrow A{M}^2=\frac{13}{4}\iff AM=\frac{\sqrt{13}}{2}$

Hướng dẫn giải:

Đáp án đúng là: D.

Theo quy tắc hình bình hành ta có:

$\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\iff {\overrightarrow{AC}}^2={\left(\overrightarrow{AB}+\overrightarrow{AD}\right)}^2$

$\iff A{C}^2=A{B}^2+2\overrightarrow{AB}.\overrightarrow{AD}+A{D}^2$

Ta có:

$cos\left(\overrightarrow{AB},\overrightarrow{AD}\right)=\cos \widehat{BAD}=cos60°=\frac{1}{2}$

$\overrightarrow{AB}.\overrightarrow{AD}=AB.AD.cos\left(\overrightarrow{AB},\overrightarrow{AD}\right)=2a.a.\frac{1}{2}=a^2$

$A{C}^2=A{B}^2+2\overrightarrow{AB}.\overrightarrow{AD}+A{D}^2=4a^2+2.a^2+a^2=7a^2$

$\Rightarrow AC=a\sqrt{7}$.

Hướng dẫn giải:

Đáp án đúng là: A.

Ta có: BI = 2CI $\Rightarrow \overrightarrow{BI}=\frac{2}{3}\overrightarrow{BC}=\frac{2}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)$

$\iff {\overrightarrow{BI}}^2=\frac{4}{9}{\left(\overrightarrow{AC}-\overrightarrow{AB}\right)}^2=\frac{4}{9}\left(A{C}^2-2\overrightarrow{AC}.\overrightarrow{AB}+A{B}^2\right)$.

Ta có:

$\cos \left(\overrightarrow{AC},\overrightarrow{AB}\right)=\cos \widehat{A}=cos{60}^o=\frac{1}{2}$

$\left|\overrightarrow{AC}\right|=AC=5;\left|\overrightarrow{AB}\right|=AB=4$

$\overrightarrow{AC}.\overrightarrow{AB}=\left|\overrightarrow{AC}\right|.\left|\overrightarrow{AB}\right|.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=\mathrm{5.4.}\frac{1}{2}=10$

$B{I}^2=\frac{4}{9}\left(A{C}^2-2\overrightarrow{AC}.\overrightarrow{AB}+A{B}^2\right)=\frac{4}{9}\left(5^2-2.10+4^2\right)=\frac{28}{3}$

$\Rightarrow BI=\frac{2\sqrt{21}}{3}$.