Three Dimensional Geometry Question 172
Question: The equation of plane through the line of intersection of planes $ ax+by+cz+d=0 $ , $ a’x+b’y+c’z+d’=0 $ and parallel to the line $ y=0,z=0 $ is
[Kurukshetra CEE 1998]
Options:
A) $ (ab’-a’b)x+(bc’-b’c)y+(ad’-a’d)=0 $
B) $ (ab’-a’b)x+(bc’-b’c)y+(ad’-a’d)z=0 $
C) $ (ab’-a’b)y+(ac’-a’c)z+(ad’-a’d)=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
The equation of a plane through the line of intersection of the planes $ ax+by+cz+d=0 $ and $ {a}‘x+{b}‘y+{c}‘z+{d}’=0 $ is $ (ax+by+cz+d)+\lambda ({a}‘x+{b}‘y+{c}‘z+{d}’)=0 $ or $ x,(a+\lambda {a}’)+y,(b+\lambda {b}’)+z,(c+\lambda {c}’)+d+\lambda {d}’=0 $ ?.(i) This is parallel to x-axis i.e., $ y=0,z=0 $
$ \therefore 1,(a+\lambda {a}’)+0,(b+\lambda {b}’)+0,(c+\lambda {c}’)=0,\Rightarrow ,\lambda =-\frac{a}{{{a}’}} $ Putting the value of l in (i), the required plane is $ y,({a}‘b-a{b}’)+z,({a}‘c-a{c}’)+{a}’d-a{d}’=0 $ i.e., $ ,(a{b^{’}}-{a^{’}}b)y+,(a{c}’-{a}‘c)z+a{d}’-{a}’d=0 $ .
BETA
