A) \[-1\]
B) \[\frac{2}{9}\]
C) \[\frac{9}{2}\]
D) 0
Correct Answer: C
Solution :
\[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\] \[\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\] \[\vec{a}\,(1,-1,1);\] \[\vec{r}\,=\vec{a}+\lambda \vec{b}\] \[\vec{b}\,(2,3,4)\] \[\vec{c}\,(3,k,0);\] \[\vec{r}=\vec{a}+\mu \vec{d}\] \[\vec{d}(1,2,1)\] These lines will intersect if lines are coplaner \[\vec{a}-\vec{c},\,\,\hat{b}\And \vec{d}\] are coplaner \[\therefore \] \[[\vec{a}-\vec{c},\vec{b},\vec{d}]=0\] \[\left| \begin{matrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow 2(-5)-(k+1)(-2)-1(1)=0\] \[\Rightarrow \,2(k+1)=11\] \[\Rightarrow k=\frac{9}{2}\]You need to login to perform this action.
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