Question

If two lines
$\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}$ and
$\frac{x-9}{2}=\frac{y-8}{4}=\frac{z-7}{8}$
lie in the same plane then the equation of that plane is -

• A. $3x-7y+z-1=0$
• B. $9x-7y+z-1=0$
• C. $6x-7y+2z-2=0$
• D. None of these

Answer 1

The correct option is C $6x-7y+2z-2=0$

We have seen that if two lines $\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$ and $\frac{x-{\alpha }^{\prime }}{l^{\prime }}=\frac{y-{\beta }^{\prime }}{m^{\prime }}=\frac{z-{\gamma }^{\prime }}{n^{\prime }}$ lie in a plane then the equation of plane is -
$\left|\begin{array}{ccc}x-\alpha & y-\beta & z-\gamma \\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$
We’ll apply the same formula to calculate the equation.
Substituting appropriate values -
$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 3& 4& 5\\ 2& 4& 8\end{array}\right|=0$
$=(x-1)12-(y-2)14+(z-3)4$
$=12x-14y+4z-4=0$
$=6x-7y+2z-2=0$