Question

(1) Determine the direction cosines of the normal to the given plane and the distance from the origin.
(i) $z=2$

(2) Determine the direction cosines of the normal to the given plane and the distance from the origin.
(ii) $x+y+z=1$

(3) Determine the direction cosines of the normal to the given plane and the distance from the origin.
(iii) $2x+3y-z=5$

(4) Determine the direction cosines of the normal to the given plane and the distance from the origin.
(iv) $5y+8=0$

Answer 1

(1) Given equation of plane is $z=2$

i.e., $0x+0y+z=2$

Comparing with $ax+by+cz=d$, we get
$a=0,b=0,c=1$ and $d=2$

Direction cosines of the normal to the plane are
$l=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{0}{\sqrt{0^2+0^2+1^2}}=0$

$m=\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{0}{\sqrt{0^2+0^2+1^2}}=0$

$n=\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{0}{\sqrt{0^2+0^2+1^2}}=1$

Direction cosines of the normal to the plane are : $0,0,1$

Distance of plane from the origin is $=\left|\frac{d}{\sqrt{a^2+b^2+c^2}}\right|$

$D=\left|\frac{2}{\sqrt{0^2+0^2+1^2}}\right|=2$

Hence, direction cosines of normal are $0,0,1$ and distance from origin is $2$

(2) Given equation of plane is $x+y+z=1$

Comparing with $ax+by+cz=d$, we get
$a=1,b=1,c=1$ and $d=1$

Direction cosines of the normal to the plane are
$l=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$

$m=\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$

$n=\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{1}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$

Direction cosines of the normal to the plane are : $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Distance of plane from the origin is $=\left|\frac{d}{\sqrt{a^2+b^2+c^2}}\right|$

$D=\left|\frac{1}{\sqrt{1^2+1^2+1^2}}\right|$

$D=\frac{1}{\sqrt{3}}$

Hence, direction cosines of normal are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$ and distance from origin is $\frac{1}{\sqrt{3}}$

(3) Given equation of plane is $2x+3y-z=5$

Comparing with $ax+by+cz=d$, we get
$a=2,b=3,c=-1$ and $d=5$

Direction cosines of the normal to the plane are
$l=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{2}{\sqrt{2^2+3^2+(-1{)}^2}}=\frac{2}{\sqrt{14}}$

$m=\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{3}{\sqrt{2^2+3^2+(-1{)}^2}}=\frac{3}{\sqrt{14}}$

$n=\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{-1}{\sqrt{2^2+3^2+(-1{)}^2}}=\frac{-1}{\sqrt{14}}$

Direction cosines of the normal to the plane are : $\frac{2}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-1}{\sqrt{14}}$

Distance of plane from the origin is ​​​​​​​$=\left|\frac{d}{\sqrt{a^2+b^2+c^2}}\right|$

$D=\left|\frac{5}{\sqrt{2^2+3^2+(-1{)}^2}}\right|$

$D=\frac{5}{\sqrt{14}}$

Hence, direction cosines of normal are $\frac{2}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-1}{\sqrt{14}}$ and distance from origin is $\frac{5}{\sqrt{14}}$

(4) Given equation of plane is $5y+8=0$

i.e., $0x+5y+0z=-8$

Comparing with $ax+by+cz=d$, we get
$a=0,b=5,c=0$ and $d=-8$

Direction cosines of the normal to the plane are
$l=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{0}{\sqrt{0^2+5^2+0^2}}=0$

$m=\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{5}{\sqrt{0^2+5^2+0^2}}=1$

$n=\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{0}{\sqrt{0^2+5^2+0^2}}=0$

Direction cosines of the normal to the plane are : $0,1,0$

Distance of plane from the origin is ​​​​​​​$=\left|\frac{d}{\sqrt{a^2+b^2+c^2}}\right|$

$D=\left|\frac{-8}{\sqrt{0^2+5^2+0^2}}\right|$

$D=\left|\frac{-8}{5}\right|=\frac{8}{5}$

Hence, direction cosines of normal are $0,1,0$ and distance from origin is $\frac{8}{5}$