+91-8076753736[email protected]
logo
logo
-->

Three Dimensional Geometry

Question
CBSEENMA12033356
Wired Faculty App

Find the normal form of the equation of the plane  3x– 4y + z + 5 = 0.

Solution

The equation of plane is 3x – 4y + z + 5 = 0
or  3 x – 4 y + z = – 5
or  – 3 x + 4 y – z = 5
Dividing both sides by square root of left parenthesis negative 3 right parenthesis squared plus left parenthesis 4 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root space equals space square root of 26 comma space space we space get
                      negative fraction numerator 3 over denominator square root of 26 end fraction straight x plus fraction numerator 4 over denominator square root of 26 end fraction straight y space minus space fraction numerator 1 over denominator square root of 26 end fraction straight z space equals space fraction numerator 5 over denominator square root of 26 end fraction
which is of form lx + my + nz = p
where  straight l equals fraction numerator negative 3 over denominator square root of 26 end fraction comma space space space straight m space equals space fraction numerator 4 over denominator square root of 26 end fraction comma space space straight n space equals negative fraction numerator 1 over denominator square root of 26 end fraction comma space space straight p space equals space fraction numerator 5 over denominator square root of 26 end fraction
The direction-cosines of the normal to the plane are fraction numerator negative 3 over denominator square root of 26 end fraction comma space fraction numerator 4 over denominator square root of 26 end fraction comma space minus fraction numerator 1 over denominator square root of 26 end fraction.

Some More Questions From Three Dimensional Geometry Chapter

If α, β, γ are the angles which a line makes with the axes, prove that sin2α + sin2 β + sin2γ = 2.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation