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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 3

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 3

Answers (1)

Answer:

 2;\: \frac{2}{7},\frac{-3}{7},\frac{-6}{7}

Hint:

 You must know the rules of finding length of perpendicular from the origin from vector equation, also find direction cosines

Given:

Reduce the equation 2x - 3y - 6z = 14 to the normal form and hence,

find the length of perpendicular from the origin to the plane. Also find direction cosines

Solution:

We have

\begin{aligned} &2x-3y-6z=14 \qquad \qquad \qquad (1)\\ &=\sqrt{(2)^{2}+(-3)^{2}+(-6)^{2}}=\sqrt{4+9+36}\\ &=\sqrt{49}=7 \end{aligned}

Divide the equation (1) with 7
We get,
\begin{aligned} &\frac{2}{7}x-\frac{3}{7}y-\frac{6}{7}z=\frac{14}{7}\\ &\frac{2}{7}x-\frac{3}{7}y-\frac{6}{7}z=2 \qquad \qquad \qquad \rightarrow (2) \end{aligned}

Now compare equation (2) with cartesian equation.
The cartesian equation of normal form of a plane,
lx + my + nz = p

where l,m,n are direction cosines and P is the length of perpendicular.
By comparing we get,
Diection cosines :
\begin{aligned} &l=\frac{2}{7},m=-\frac{3}{7}, n=-\frac{6}{7} \end{aligned}

Length of perpendicular from the origin P = 2

Posted by

Gurleen Kaur

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