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
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., \[\,(a{{b}^{'}}-{{a}^{'}}b)y+\,(a{c}'-{a}'c)z+a{d}'-{a}'d=0\].
You need to login to perform this action.
You will be redirected in
3 sec
