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Home » In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane (a) (2, 3, -5) x + 2y – 2z = 9 (b) (-6, 0, 0) 2x – 3y + 6z – 2 = 0
In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane
(a) (2, 3, -5) x + 2y – 2z = 9
(b) (-6, 0, 0) 2x – 3y + 6z – 2 = 0
CBSE | Class 12 | Exercise 11.3 | Maths | NCERT | Three Dimensional Geometry
In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane
(a) (2, 3, -5) x + 2y – 2z = 9
(b) (-6, 0, 0) 2x – 3y + 6z – 2 = 0

Solution:

(a) The length of perpendicular from the point (2,3,-5) on the plane x+2 y-2 z=9 \Rightarrow x+2 y-2 z-9=0 is
\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}
=\frac{|2(1)+2(3)-2(-5)-9|}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}
=\frac{|9|}{\sqrt{1+4+4}}=\frac{9}{\sqrt{9}}=\frac{9}{3}=3

(b) The length of perpendicular from the point (-6,0,0) on the plane 2 x-3 y+6 z-2=0 is
\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}
\begin{aligned} \frac{|2(-6)-3(0)+6(0)-2|}{\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}} \\ = \frac{|-14|}{\sqrt{4+9+36}}=\frac{14}{\sqrt{49}}=\frac{14}{7}=2 \end{aligned}

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