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

$\begin{array}{1 1} (a)\:\:ax_1+by_1+cz_1=0 & (b)\:\:al+bm+cn=0 \\ (c)\:\:\large\frac{a}{l}=\frac{b}{m}=\frac{c}{n} & (d)\:\:x_1+my_1+nz_1=0\end{array}$
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## 1 Answer

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Toolbox:
• Every line on a plane is $\perp$ to the normal to the plane.
• If two lines are $\perp$, then dot product of their $d.r.$ =$0$.
Given line $\large\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$..........(i)
$\Rightarrow\:d.r.$ of the line, $\overrightarrow b$ $= (l,m,n)$
Plane $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
$\therefore\:$ $d.r.$ of the normal, $\overrightarrow n=(a,b,c)$
If the line lies on the plane then every point on the line satisfies the equation of the plane and
the line is $\perp$ to the normal to the plane.
$\Rightarrow\:\overrightarrow n.\overrightarrow b=0$
$\Rightarrow\:(a,b,c).(l,m,n)=0$ or $al+bn+cn=0$
answered Dec 29, 2013

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