Trong không gian Oxyz, cho tứ diện ABCD với A(3; −1; 1), B(−1; 0; 0), C(0; 1; 0), D(0; 0; 2). Chiều cao AH

Đáp án đúng là: D

Với A(3; −1; 1), B(−1; 0; 0), C(0; 1; 0), D(0; 0; 2) ta có:

• $\overrightarrow{BA}$ = (4; −1; 1);

• $\overrightarrow{BC}$ = (1; 1; 0);

• $\overrightarrow{B\text{D}}$ = (1; 0; 2);

• $\left[\overrightarrow{BC},\overrightarrow{B\text{D}}\right]$ = (1.2 − 0.0; 0.1 − 1.2; 1.0 − 1.1)

Þ $\left[\overrightarrow{BC},\overrightarrow{B\text{D}}\right]$ = (2; −2; −1);

 • $\left[\overrightarrow{BC},\overrightarrow{B\text{D}}\right]$.$\overrightarrow{BA}$= 4.2 + (−1). (−2) + 1. (−1) = 9

V_ABCD = $\frac{1}{6}$.$\left[\overrightarrow{BC},\overrightarrow{B\text{D}}\right]$.$\overrightarrow{BA}$ = $\frac{1}{6}$.9 = $\frac{3}{2}$ (đvtt)

S_BCD = $\frac{1}{2}$.$\left|\left[\overrightarrow{BC},\overrightarrow{B\text{D}}\right]\right|$ = $\frac{1}{2}$.$\sqrt{2^2+{\left(-2\right)}^2+{\left(-1\right)}^2}$ = $\frac{3}{2}$ (đvdt)

Mặt khác: V_ABCD = $\frac{1}{3}$.AH. S_BCD

Þ AH = $\frac{3V_{ABC\text{D}}}{S_{BC\text{D}}}$= $\frac{3.\frac{3}{2}}{\frac{3}{2}}$ = 3.

Vậy ta chọn phương án D.