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\(\overrightarrow{u}\) ⊥\(\overrightarrow{v}\)
=>\(\overrightarrow{u}\cdot\overrightarrow{v}=0\)
=>\(\left(\overrightarrow{a}-3\cdot\overrightarrow{b}\right)\left(2\cdot\overrightarrow{a}+\overrightarrow{b}\right)=0\)
=>\(2\cdot\left(\overrightarrow{a}\right)^2+\overrightarrow{a}\cdot\overrightarrow{b}-6\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot\left(\overrightarrow{b}\right)^2=0\)
=>\(2\cdot2-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}-3\cdot3=0\)
=>\(-5\cdot\overrightarrow{a}\cdot\overrightarrow{b}=5\)
=>\(\overrightarrow{a}\cdot\overrightarrow{b}=-1\)
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.
\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)
bài 1
a CO-OB=BA
<=.> CO = BA +OB
<=> CO=OA ( LUÔN ĐÚNG )=>ĐPCM
b AB-BC=DB
<=> AB=DB+BC
<=> AB=DC(LUÔN ĐÚNG )=> ĐPCM
Cc DA-DB=OD-OC
<=> DA+BD= OD+CO
<=> BA= CD (LUÔN ĐÚNG )=> ĐPCM
d DA-DB+DC=0
VT= DA +BD+DC
= BA+DC
Mà BA=CD(CMT)
=> VT= CD+DC=O
Vì \(\overrightarrow{a}\) ⊥\(\overset{}{\overrightarrow{b}}\) nên ta có: \(\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{0}\)
=>\(\left|\overrightarrow{a}-\overrightarrow{b}\right|=\left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{\left(\left|\overrightarrow{a}\right|\right)^2+\left(\left|\overrightarrow{b}\right|\right)^2}\)
\(=\sqrt{5^2+12^2}=13\)