Question

The equation of the plane containing the line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0,$ then which of the following is true?

• A. $a{x}_1+b{y}_1+c{z}_1=0$
• B. $al+bm+cn=0$
• C. $\frac{a}{l}=\frac{b}{m}=\frac{c}{n}$
• D. $l{x}_1+m{y}_1+n{z}_1=0$

Answer 1

The correct option is B $al+bm+cn=0$

If a plane contains a line, then the normal of the plane must be perpendicular to the line.
$D.R^{\prime }s$ of the line are $(l,m,n)$
$D.R^{\prime }s$ of normal of the plane are $(a,b,c)$
For both to be perpendicular $al+bm+cn=0$