Cho hình vuông \[ABCD\] cạnh \[a\]. Tính \(P = \overrightarrow {AC} .\left( {\overrightarrow {CD} + \overrightarrow {CA} } \right)\)

a) Tính \(\overrightarrow {AB}  \cdot \overrightarrow {BD} \). Ta có: \(\overrightarrow {AB}  \cdot \overrightarrow {BD}  = \overrightarrow {AB} (\overrightarrow {BA}  + \overrightarrow {AD} ) = \overrightarrow {AB}  \cdot \overrightarrow {BA}  + \underbrace {\overrightarrow {AB}  \cdot \overrightarrow {AD} }_0\)

\( = \overrightarrow {AB}  \cdot \overrightarrow {BA}  =  - {\overrightarrow {AB} ^2} =  - A{B^2} =  - 4{a^2}{\rm{. }}\)

b) Tính \(\overrightarrow {BC}  \cdot \overrightarrow {BD} \). Ta có: \(\overrightarrow {BC}  \cdot \overrightarrow {BD}  = BC \cdot BD \cdot \cos (\overrightarrow {BC} ,\overrightarrow {BD} ) = BC \cdot BD \cdot \cos \widehat {DBC}\)

\( = BC \cdot BD \cdot \cos \widehat {BDA} = BC \cdot BD \cdot \frac{{AD}}{{BD}} = BC \cdot AD = 3{a^3}{\rm{. }}\)

(trong đó \(\widehat {DBC} = \widehat {BDA}\) vì là hai góc so le trong).

c) Tính \(\overrightarrow {AC}  \cdot \overrightarrow {BD} \).

Ta có: \(\overrightarrow {AC}  \cdot \overrightarrow {BD}  = (\overrightarrow {AB}  + \overrightarrow {BC} )(\overrightarrow {BA}  + \overrightarrow {AD} ) = \overrightarrow {AB}  \cdot \overrightarrow {BA}  + \overrightarrow {AB}  \cdot \overrightarrow {AD}  + \overrightarrow {BC}  \cdot \overrightarrow {BA}  + \overrightarrow {BC}  \cdot \overrightarrow {AD} \)

\( =  - {\overrightarrow {AB} ^2} + 0 + 0 + BC \cdot AD \cdot \cos {0^0} =  - A{B^2} + 3a \cdot a \cdot 1 =  - {(2a)^2} + 3{a^2} =  - {a^2}.\)

d) Tính \(\overrightarrow {AC}  \cdot \overrightarrow {IJ} \). Ta có:

\(\overrightarrow {AC}  \cdot \overrightarrow {IJ}  = (\overrightarrow {AB}  + \overrightarrow {BC} ) \cdot \overrightarrow {IJ}  = \underbrace {\overrightarrow {AB}  \cdot \overrightarrow {IJ} }_0 + \overrightarrow {BC}  \cdot \overrightarrow {IJ}  = BC \cdot IJ \cdot \cos {0^0} = 3a \cdot 2a \cdot 1 = 6{a^2}.\)

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{\overrightarrow {IJ}  = \overrightarrow {IA}  + \overrightarrow {AC}  + \overrightarrow {CJ} }\\{\overrightarrow {IJ}  = \overrightarrow {ID}  + \overrightarrow {DB}  + \overrightarrow {BJ} }\end{array} \Rightarrow 2\overrightarrow {IJ}  = \overrightarrow {AC}  + \overrightarrow {DB} } \right.\).

Suy ra: \(\overrightarrow {HK}  \cdot 2\overrightarrow {IJ}  = \overrightarrow {HK} (\overrightarrow {AC}  + \overrightarrow {DB} ) = \overrightarrow {HK}  \cdot \overrightarrow {AC}  + \overrightarrow {HK}  \cdot \overrightarrow {DB} \)

\( = (\overrightarrow {HB}  + \overrightarrow {BD}  + \overrightarrow {DK} )\overrightarrow {AC}  + (\overrightarrow {HA}  + \overrightarrow {AC}  + \overrightarrow {CK} )\overrightarrow {DB}  = \overrightarrow {AC} (\overrightarrow {BD}  + \overrightarrow {DB} ) = \overrightarrow {AC}  \cdot \vec 0 = 0\).

Vậy \(\overrightarrow {HK}  \cdot \overrightarrow {IJ}  = 0\)