Câu hỏi: Cho hình hộp chữ nhật ABCD.A'B'C'D' có $AB=a,AD=a\sqrt{3}.$ Khoảng cách giữa hai đường thẳng DD' và AC' bằng
A. $\dfrac{a\sqrt{3}}{4}.$
B. $a\sqrt{3}.$
C. $\dfrac{a\sqrt{3}}{2}.$
D. $\dfrac{a\sqrt{2}}{2}.$
Ta có $d\left( D{D}',A{C}' \right)=d\left( B{B}',A{C}' \right).$
Ta có ${A}'{C}'=\sqrt{{{\left( {A}'{B}' \right)}^{2}}+{{\left( {B}'{C}' \right)}^{2}}}=2a.$
Kẻ ${B}'H\bot {A}'{C}'.$
${B}'H=\dfrac{{A}'{B}'.{B}'{C}'}{{A}'{C}'}=\dfrac{a.a\sqrt{3}}{2a}=\dfrac{a\sqrt{3}}{2}.$
Vì $B{B}'//\left( AC{C}'{A}' \right)$ nên $d\left( B{B}',A{C}' \right)=d\left( B{B}',\left( AC{C}'{A}' \right) \right)$
$d\left( B{B}',\left( AC{C}'{A}' \right) \right)={B}'H=\dfrac{a\sqrt{3}}{2}.$
Nên $d\left( B{B}',A{C}' \right)=\dfrac{a\sqrt{3}}{2}.$
A. $\dfrac{a\sqrt{3}}{4}.$
B. $a\sqrt{3}.$
C. $\dfrac{a\sqrt{3}}{2}.$
D. $\dfrac{a\sqrt{2}}{2}.$
Ta có $d\left( D{D}',A{C}' \right)=d\left( B{B}',A{C}' \right).$
Ta có ${A}'{C}'=\sqrt{{{\left( {A}'{B}' \right)}^{2}}+{{\left( {B}'{C}' \right)}^{2}}}=2a.$
Kẻ ${B}'H\bot {A}'{C}'.$
${B}'H=\dfrac{{A}'{B}'.{B}'{C}'}{{A}'{C}'}=\dfrac{a.a\sqrt{3}}{2a}=\dfrac{a\sqrt{3}}{2}.$
Vì $B{B}'//\left( AC{C}'{A}' \right)$ nên $d\left( B{B}',A{C}' \right)=d\left( B{B}',\left( AC{C}'{A}' \right) \right)$
$d\left( B{B}',\left( AC{C}'{A}' \right) \right)={B}'H=\dfrac{a\sqrt{3}}{2}.$
Nên $d\left( B{B}',A{C}' \right)=\dfrac{a\sqrt{3}}{2}.$
Đáp án C.