Question

Find the value of $a$ for which the lines $x-2=\frac{y-9}{2}=\frac{z-13}{3}$ and $\frac{z-a}{-1}=\frac{y-7}{2}+\frac{z+2}{-3}$ intersect :

• A. $5$
• B. $-3$
• C. $4$
• D. $Noneofthese$

Answer 1

The correct option is B $-3$

$\frac{x-2}{1}=\frac{y-9}{2}=\frac{2-13}{3},\frac{x-1}{-1}=\frac{y-7}{2}=\frac{z+2}{-3}$

Condition for two lines intersect is

$\begin{array}{c}\left(\begin{array}{ccc}{x}_2-x_1& y{}_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right)=0\\ where=\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{b_2}....\\ \left(\begin{array}{ccc}\left(-a+x\right)& -7+9& 2+13\\ 1& 2& 3\\ -1& 2& -3\end{array}\right)=0\\ Solving\\ \left(-a+2\right)\left(-6-6\right)-2\left(-3+3\right)+15\left(2+2\right)\\ 12a-24+60=0\\ 12a=-36\\ a=\frac{-36}{12}=a=-3\end{array}$