Question

The condition that the line $\frac{x-{\alpha }^{\prime }}{l}=\frac{y-{\beta }^{\prime }}{m}=\frac{z-{\gamma }^{\prime }}{n}$ in the plane $Ax+By+Cz+D=0$ is

• A. $A{\alpha }^{\prime }+B{\beta }^{\prime }+C{\gamma }^{\prime }+D=0andAl+Bm+Cn\ne 0$
• B. $A{\alpha }^{\prime }+B{\beta }^{\prime }+C{\gamma }^{\prime }+D\ne 0andAl+Bm+Cn=0$
• C. $A{\alpha }^{\prime }+B{\beta }^{\prime }+C{\gamma }^{\prime }+D=0andAl+Bm+Cn=0$
• D. $A{\alpha }^{\prime }+B{\beta }^{\prime }+C{\gamma }^{\prime }=0andAl+Bm+Cn=0$

Answer 1

The correct option is C $A{\alpha }^{\prime }+B{\beta }^{\prime }+C{\gamma }^{\prime }+D=0andAl+Bm+Cn=0$

The line $\frac{x-{\alpha }^1}{l}=\frac{y-{\beta }^1}{m}=\frac{z-{\gamma }^1}{n}$ in the plane $Ax+By+Cz+D=0$

then multiplication sum id corresponding direction ratio will be zero so here $Al+Bm+Cn=0$

and one thing more line passes through $({\alpha }^1,{\beta }^1,{\gamma }^1)$

So, plane Also passe through $({\alpha }^1,{\beta }^1,{\gamma }^1)$

hence $A{\alpha }^1+B{\beta }^1+C{\gamma }^1+D=0$