Đáp án:
\(\left( {\overrightarrow {IA} + \overrightarrow {IB} } \right).\overrightarrow {ID} = - \frac{{9{a^2}}}{2}.\)
Giải thích các bước giải:
Xét \(\Delta ABI\) vuông tại \(A\) ta có:
\(\begin{array}{l}IB = \sqrt {I{A^2} + A{B^2}} = \sqrt {{{\left( {\frac{{3a}}{2}} \right)}^2} + {{\left( {4a} \right)}^2}} = \frac{{a\sqrt {73} }}{2}.\\ \Rightarrow \cos \angle AIB = \frac{{IA}}{{IB}} = \frac{{3a}}{2}.\frac{2}{{a\sqrt {73} }} = \frac{{3\sqrt {73} }}{{73}}.\end{array}\)
Mà \(\angle AIB\) và \(\angle DIB\) là hai góc kề bù
\( \Rightarrow \cos \angle DIB = - \cos \angle AIB = - \frac{{3\sqrt {73} }}{{73}}.\)
Ta có:
\(\begin{array}{l}\left( {\overrightarrow {IA} + \overrightarrow {IB} } \right).\overrightarrow {ID} = \overrightarrow {IA} .\overrightarrow {ID} + \overrightarrow {IB} .\overrightarrow {ID} \\ = IA.ID.\cos \left( {\overrightarrow {IA} ,\,\,\overrightarrow {ID} } \right) + IB.ID.\cos \left( {\overrightarrow {IB} ,\,\,\overrightarrow {ID} } \right)\\ = IA.ID.\cos {180^0} + IB.ID.\cos \angle DIB\\ = \frac{{3a}}{2}.\frac{{3a}}{2}.\left( { - 1} \right) + \frac{{3a}}{2}.\frac{{a\sqrt {73} }}{2}\left( { - \frac{{3\sqrt {73} }}{{73}}} \right)\\ = - \frac{{9{a^2}}}{4} - \frac{{9{a^2}}}{4}\\ = - \frac{{9{a^2}}}{2}.\end{array}\)
