Question

The perpendicular distance of the origin form the plane which makes intercepts 12, 3 and 4 on x, y, z axes respectively, is

• A. 13
• B. 11
• C. 17
• D. $\frac{6\sqrt{2}}{\sqrt{1}3}$

Answer 1

The correct option is D $\frac{6\sqrt{2}}{\sqrt{1}3}$

Let equation of plane be lx + my + nz = p, where p is the perpendicular distance of the plane from origin and (l,m,n) are the direction cosines of the normal to the plane
Or $\frac{x}{\left(\frac{p}{l}\right)}+\frac{y}{\left(\frac{p}{m}\right)}+\frac{z}{\left(\frac{p}{n}\right)}=1$
According to question,
$\frac{p}{l}=12,\frac{p}{m}=3,\frac{p}{n}=4$
or $\frac{p}{12}=l,\frac{p}{3}=m,\frac{p}{4}=n$
or $\frac{p^2}{144}+\frac{p^2}{9}+\frac{p^2}{16}=l^2+m^2+n^2=1$
$\Rightarrow p^2(\frac{1}{144}+\frac{1}{9}+\frac{1}{16})=1p^2\left(\frac{1+16+9}{144}\right)=1$
Or $p^2=\frac{144}{26}=\frac{72}{13}$
$\therefore p=\frac{6\sqrt{2}}{\sqrt{13}}$