Question

The equation of the line passing through the point (1,2,–4) and perpendicular to the two lines $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$,

• A. $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$
  
• B. $\frac{x-1}{-2}=\frac{y-2}{3}=\frac{z+4}{8}$
  
• C. $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+4}{8}$
• D. $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{4}$

Answer 1

The correct option is A $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$


Line passing through the point (1,2,-4) is $\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+4}{n}$
Now, according to question, 3l -16m +7n =0 and 3l+8m -5n =0
by solving (l,m,n) = (2,3,6)
Hence required line is, $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$