Trong không gian \(Oxyz\), cho ba đường thẳng \(d:\frac{x}{1}=\frac{y}{1}=\frac{z+1}{-2},\)\({{\Delta }_{1}}:\frac{x-3}{2}=\frac{y}{1}=\frac{z-1}{1},\)\({{\Delta }_{2}}:\frac{x-1}{...

* Hướng dẫn giải

\(H\in {{\Delta }_{1}}\Leftrightarrow H\left( 3+2t;t;1+t \right)\), \(K\in {{\Delta }_{2}}\Leftrightarrow K\left( 1+m;2+2m;m \right)\).

Ta có\(\overrightarrow{HK}=\left( m-2t-2;2m-t+2;m-t-1 \right)\). Đường thẳng \(d\) có một VTCP là \(\overrightarrow{{{u}_{d}}}=\left( 1;1;-2 \right)\).

\(\Delta \bot d\Leftrightarrow \)\(\overrightarrow{{{u}_{d}}}.\overrightarrow{HK}=0\)\(\Leftrightarrow m-t+2=0\Leftrightarrow m=t-2\Rightarrow \overrightarrow{HK}=\left( -t-4;t-2;-3 \right).\)

Ta có\(H{{K}^{2}}={{\left( -t-4 \right)}^{2}}+{{\left( t-2 \right)}^{2}}+{{\left( -3 \right)}^{2}}=2{{\left( t+1 \right)}^{2}}+27\ge 27,\forall t\in \mathbb{R}\).

\(HK=\sqrt{27}\Leftrightarrow t=-1,\,\,m=-3.\) Khi đó \(\overrightarrow{HK}=\left( -3;-3;-3 \right)=-3(1;1;1)\), \(H(1;-1;0)\).

Phương trình đường thẳng \(\Delta \) là \(\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{1}\).