Ta có  a → = m . u → + v → = 4 m + 1 ; m + 4 b → = i → + j → = 1 ; 1 .

Yêu cầu bài toán  ⇔ cos a → , b → = cos 45 0 = 2 2

⇔ 4 m + 1 .1 + m + 4 .1 2 4 m + 1 2 + m + 4 2 = 2 2 ⇔ 5 m + 1 2 17 m 2 + 16 m + 17 = 2 2

⇔ 5 m + 1 = 17 m 2 + 16 m + 17 ⇔ m + 1 ≥ 0 25 m 2 + 50 m + 25 = 17 m 2 + 16 m + 17 ⇔ m = − 1 4 .

Chọn C.

u(1/2;-5).    v(k;-4)

\(\overrightarrow{v}=3\overrightarrow{i}-m\overrightarrow{j}\Rightarrow\overrightarrow{v}=\left(3;-m\right)\)

Để \(\overrightarrow{u};\overrightarrow{v}\) cùng phương:

\(\Leftrightarrow\frac{3}{-2}=\frac{-m}{1}\Rightarrow m=\frac{3}{2}\)

\(\overrightarrow{v}=\left(3;-m\right)\)

Hai vecto đã cho cùng phương khi và chỉ khi:

\(\dfrac{3}{-2}=\dfrac{-m}{1}\Leftrightarrow m=\dfrac{3}{2}\)

\(\overrightarrow{c}=2\left(2;1\right)+3\left(3;-2\right)=\left(4+9;2-6\right)=\left(13;-4\right)\)

a.

\(\overrightarrow{u}=2\left(2;1\right)-\left(3;4\right)=\left(1;-2\right)\)

\(\overrightarrow{v}=3\left(3;4\right)-2\left(7;2\right)=\left(-5;8\right)\)

\(\overrightarrow{w}=5\left(7;2\right)+\left(2;1\right)=\left(37;11\right)\)

b.

\(\overrightarrow{x}=2\left(2;1\right)+\left(3;4\right)-\left(7;2\right)=\left(0;4\right)\)

\(\overrightarrow{z}=2\left(2;1\right)-3\left(3;4\right)+\left(7;2\right)=\left(2;-8\right)\)

c.

\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)

\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)

a) Vì \(\overrightarrow v  = \left( {0; - 7} \right)\)nên \(\overrightarrow v  = 0\overrightarrow i  + \left( { - 7} \right)\overrightarrow j  =  - 7\overrightarrow j \)

b) Vì B có tọa  độ là (-1; 0) nên \(\overrightarrow {OB}  = \left( { - 1;{\rm{ }}0} \right)\). Do đó: \(\overrightarrow {OB}  = \left( { - 1} \right)\overrightarrow i  + 0\overrightarrow j  =  - \overrightarrow i \)

\(\overrightarrow{a}=\left(2;-1\right)\)