Trong không gian Oxyz, cho hai điểm A(1;3;0); B(4;3;3) và đường thẳng d x+ 5/5 = y +3/4 = z/1

Đáp án B

Do $M\in d\Rightarrow M\left(5t-5;4t-3;t\right)$

$\Rightarrow \left\{\begin{array}{l}\overrightarrow{AM}=\left(5t-6;4t-6;t\right)\\ \overrightarrow{BM}=\left(5t-9;4t-6;t-3\right)\end{array}\right.\Rightarrow \left\{\begin{array}{l}M{A}^2=42t^2-108t+72\\ M{B}^2=42t^2-144t+126\\ M{A}^2+M{B}^2=84t^2-252t+198\left(*\right)\end{array}\right.$

Ta có: $A{B}^2+M{A}^2+M{B}^2-2MA.MB.\cos \widehat{AMB}$

$\begin{array}{l}\iff 18=84t^2-252t+198-\sqrt{\left(42t^2-108t+72\right)\left(42t^2-144t+126\right)}\\ \iff 84t^2-252t+180-\sqrt{\left(42t^2-108t+72\right)\left(42t^2-144t+126\right)}=0\stackrel{Casio}{\to }t=2\end{array}$

Thay t = 2 vào (*), ta được T = 30.