# NCERT Solutions Class 12 Maths Chapter-11 (Three Dimensional Geometry)Exercise 11.3

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-11 (Three Dimensional Geometry)Exercise 11.3 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

### Exercise 11.3

Q1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
Answer. $\begin{array}{}\end{array}$ (a) The equation of the plane is z=2 or 0x+0y+z=2 (1)  The direction ratios of normal are 0,0, and 1 . 02+02+12=1    This equation [s of the form + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance Of normal from the origin. Therefore, the direction cosines of the normal are  and the distance of normal form the origin is $\frac{1}{\sqrt{3}}$ unints.  $\frac{2}{\sqrt{14}}x+\frac{3}{\sqrt{14}}y-\frac{1}{\sqrt{14}}z=\frac{5}{\sqrt{14}}$ This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. Therefore, the direction cosines of the normal to the plane are  and the distance of normal from the origin is $\frac{5}{\sqrt{14}}$ units.  The direction ratios of normal are 0, -5, and 0.  This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. Therefore, the direction cosines of the normal to the plane are 0, -1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.

Q2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\stackrel{^}{i}+5\stackrel{^}{ȷ}-6\stackrel{^}{k}$.

Q3. Find the Cartesian equation of the following planes: (a)  (c)

Answer.   This is the cartesian equation of the plane.  Substituting the value of in equation (1), we obtain   Substituting the value of in equation (1), we obtain $\begin{array}{l}\left(x\stackrel{^}{i}+y\stackrel{^}{y}-z\stackrel{^}{k}\right)\cdot \left[\left(s-2t\right)\stackrel{^}{i}+\left(3-t\right)\stackrel{^}{j}+\left(2s+t\right)\stackrel{^}{k}\right]=15\\ ⇒\left(s-2t\right)x+\left(3-t\right)y+\left(2s+t\right)z=15\end{array}$ This is the cartesian equation of the given plane.

Q4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

Answer. (a) Let the coordinates of the foot of perpendicular P from the origin to the plane be $\left({x}_{1},{y}_{1},{z}_{1}\right)$ $\begin{array}{l}2x+3y+4z-12=0\\ ⇒2x+3y+4z=12\dots \left(1\right)\end{array}$ The direction ratios of normal are 2, 3, and 4.  This equation of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. The coordinates of the foot of the perpendicular are given by (ld, md, nd) Therefore, the coordinates of the foot of the perpendicular are  (b) Let the coordinates of the foot of perpendicular P from the origin to the plane be $\left({x}_{1},{y}_{1},{z}_{1}\right)$$\begin{array}{l}3y+4z-6=0\\ ⇒\phantom{\rule{1em}{0ex}}0x+3y+4z=6\end{array}$ The direction ratios of the normal are 0, 3, and 4.  This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. The coordinates of the foot of the perpendicular are given by (ld, md, nd). Therefore, the coordinates of the foot of the perpendicular are  (c) Let the coordinates of the foot of perpendicular P from the origin to the plane be $\left({x}_{1},{y}_{1},{z}_{1}\right)$. X + y + z = 1 The direction ratios of the normal are 1, 1 and 1. $\therefore \sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3}$ Dividing both sides of equation (1) by $3$ , we obtain $\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}$ This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. The coordinates of the foot of the perpendicular are given by (ld, md, nd).  (d) Let the coordinates of the foot of perpendicular P from the origin to the plane be $\left({x}_{1},{y}_{1},{z}_{1}\right)$. 5y + 8 = 0 $⇒0x-5y+0z=8\dots \left(1\right)$ The direction ratios of the normal are 0, -5, and 0.  This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. The coordinates of the foot of the perpendicular are given by (ld, md, nd ). Therefore, the coordinates of the foot of the perpendicular are

Q5. Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is $\stackrel{^}{\stackrel{^}{i}}+\stackrel{^}{j}-\stackrel{^}{k}$. (b) that passes through the point (1,4, 6) and the normal vector to the plane is $\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}$.

Answer.  The vector equation of the plane is given by, $\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right)\stackrel{\to }{N}=0$ $⇒\left[\stackrel{\to }{r}-\left(i-2\stackrel{^}{k}\right)\right]\cdot \left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)=0$ $\stackrel{\to }{r}$ is the position vector of any point P (x, y, z) in the plane. $\therefore \stackrel{\to }{r}=x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$ Therefore, equation (1) becomes