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 (where $a,b,c$ are direction ratios normal to plane)

• 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. $l{x}_1+m{y}_1+n_1=0$

Answer 1

The correct option is B

$al+bm+cn=0$

Explanation for correct option:

The equation of the plane containing the line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ is

$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$ where $l,m,n$ are direction ratios of line and $a,b,c$ are direction ratios normal to plane.

Since, the normal to the plane ($a,b,c$) is perpendicular to the given line ($l,m,n$).

Therefore, $al+bm+cn=0$

Hence option(B) is correct.