Question

The equation of the plane through the line of intersection of the planes $ax+by+cz+d=0$ and $a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime }=0$ and parallel to the line $y=0$ and $z=0$ is-

• A. $(a{b}^{\prime }-a^{\prime }b)x+(b{c}^{\prime }-b^{\prime }c)y+(a{d}^{\prime }-a^{\prime }d)=0$
• B. $(a{b}^{\prime }-a^{\prime }b)x+(b{c}^{\prime }-b^{\prime }c)y+(a{d}^{\prime }-a^{\prime }d)z=0$
• C. $(a{b}^{\prime }-a^{\prime }b)y+(a{c}^{\prime }-a{c}^{\prime })z=(a{d}^{\prime }-a^{\prime }d)$
• D. None of these.

Answer 1

Answer: (D)

None of these.

The equation of any plane through the line of intersection of the given plane is
$ax+by+cz+d+\lambda (a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })=0$
$\Rightarrow x(a+a^{\prime }\lambda )+y(b+b^{\prime }\lambda )+z(c+c^{\prime }\lambda )+(d+d^{\prime }\lambda )=0$ ...(2)
Now the d.c's of $x$-axis are $1,0,0$ and the d.r's of the normal to plane (2) are $a+a^{\prime }\lambda ,b+b^{\prime }\lambda ,c+c^{\prime }\lambda$
The plane (2) is parallel to $x$-axis of the normal to the plane (2) is perpendicular to the $x$-axis, the condition for which is
$1.(a+a^{\prime }\lambda )+0.(b+b^{\prime }\lambda )+0.(c+c^{\prime }\lambda )=0$ giving $\lambda =-\frac{a}{a^{\prime }}$
Putting this value of $\lambda$ in the equation (1), the required equation of the plane is given by
$a^{\prime }(ax+by+cz+d)-a(a^{\prime }x+b^{\prime }y+c^{\prime }z+d^{\prime })\Rightarrow (b{a}^{\prime }-a{b}^{\prime })y+(c{a}^{\prime }-a{c}^{\prime })z+(d{a}^{\prime }-a{d}^{\prime })=0$