Cho hình lập phương ABCD.ABCD có cạnh bằng a. Gọi K là trung điểm của DD". Tìm khoảng cách giữa CK, A'D.
Đồng nhất hình vẽ với hệ trục toạn độ trong không gian sao cho A' \XiO, A'B' \XiOx, A'D'\Xi Oy, A'A\XiOz
Khi đó, tọa độ các điểm là: A'(0,0,0), D(0,a,a), C( a,a,a), K(0,a,a/2)
Theo công thức tính khoảng cách 2 đường chéo nhau trong không gian tọa dộ. Ta có:
d( A'D, CK)=\frac{\left| {\left[ {\overrightarrow{A'D},\overrightarrow{CK}} \right].\overrightarrow{KD}} \right|}{\left| {\left[ {\overrightarrow{A'D},\overrightarrow{CK}} \right]} \right|}
Mà \overrightarrow{A'D}=(0,a,a); \overrightarrow{CK}=( -a,0,-a/2); \overrightarrow{KD}=(0,0,-a/2)
=> d(A'D, CK)=\frac{a}{3}

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