NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.2) Exercise 12.2

Exercise 12.2

1. Find the distance between the following pairs of points:

(i). \[\left( {2,3,5} \right)\] and $\left( {4,3,1} \right)$

Ans: We have, distance between two points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ .

So, the distance between the point \[{\text{P}}\left( {2,3,5} \right)\] and ${\text{Q}}\left( {4,3,1} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {3 - 3} \right)}^2} + {{\left( {1 - 5} \right)}^2}} $

${\text{PQ}} = \sqrt {{2^2} + {0^2} + {{\left( { - 4} \right)}^2}} $

${\text{PQ}} = \sqrt {4 + 0 + 16} $

${\text{PQ}} = \sqrt {20} $

By square root,

${\text{PQ}} = 2\sqrt 5 $units

Hence, the distance between the pairs of points \[\left( {2,3,5} \right)\] and $\left( {4,3,1} \right)$ is $2\sqrt 5 $ units.

(ii). \[\left( { - 3,7,2} \right)\] and $\left( {2,4, - 1} \right)$

Ans: We have, distance between two points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ .

So, the distance between the point \[{\text{P}}\left( { - 3,7,2} \right)\] and ${\text{Q}}\left( {2,4, - 1} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {2 - \left( { - 3} \right)} \right)}^2} + {{\left( {4 - 7} \right)}^2} + {{\left( { - 1 - 2} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {4 - 7} \right)}^2} + {{\left( { - 1 - 2} \right)}^2}} $

${\text{PQ}} = \sqrt {{5^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 3} \right)}^2}} $

${\text{PQ}} = \sqrt {43} $units

Hence, the distance between the pairs of points \[\left( { - 3,7,2} \right)\] and $\left( {2,4, - 1} \right)$is$\sqrt {43} $ units.

(iii). \[\left( { - 1,3, - 4} \right)\] and $\left( {1, - 3,4} \right)$

Ans: We have, distance between two points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ .

So, the distance between the point \[{\text{P}}\left( { - 1,3, - 4} \right)\] and ${\text{Q}}\left( {1, - 3,4} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left( { - 3 - 3} \right)}^2} + {{\left( {4 - \left( { - 4} \right)} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( {1 + 1} \right)}^2} + {{\left( { - 3 - 3} \right)}^2} + {{\left( {4 + 4} \right)}^2}} $

${\text{PQ}} = \sqrt {{2^2} + {{\left( { - 6} \right)}^2} + {{\left( 8 \right)}^2}} $

${\text{PQ}} = \sqrt {4 + 36 + 64} $

${\text{PQ}} = \sqrt {104} $

By square root,

${\text{PQ}} = 2\sqrt {26} $units

Hence, the distance between the pairs of points \[\left( { - 1,3, - 4} \right)\] and $\left( {1, - 3,4} \right)$is$2\sqrt {26} $ units.

(iv). \[\left( {2, - 1,3} \right)\] and $\left( { - 2,1,3} \right)$

Ans: We have, distance between two points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$ is

${\text{PQ}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ .

So, the distance between the point \[{\text{P}}\left( {2, - 1,3} \right)\] and ${\text{Q}}\left( { - 2,1,3} \right)$ is

${\text{PQ}} = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left( {3 - 3} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {1 + 1} \right)}^2} + {{\left( {3 - 3} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( { - 4} \right)}^2} + {2^2} + {0^2}} $

${\text{PQ}} = \sqrt {16 + 4 + 0} $

${\text{PQ}} = \sqrt {20} $

By square root,

${\text{PQ}} = 2\sqrt 5 $units

Hence, the distance between the pairs of points \[\left( {2, - 1,3} \right)\] and $\left( { - 2,1,3} \right)$is$2\sqrt 5 $ units.

2. Show that the points $\left( { - 2,3,5} \right),\left( {1,2,3} \right)$ and $\left( {7,0, - 1} \right)$ are collinear.

Ans: Let the points are ${\text{P}}\left( { - 2,3,5} \right),{\text{Q}}\left( {1,2,3} \right)$and ${\text{R}}\left( {7,0, - 1} \right)$.

We have, the points are said to be collinear if they lie on a line.

So, we have to show that ${\text{PQ}} + {\text{QR}} = {\text{PR}}$.

Now, ${\text{PQ}} = \sqrt {{{\left( {1 - \left( { - 2} \right)} \right)}^2} + {{\left( {2 - 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( {2 - 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}} $

${\text{PQ}} = \sqrt {{3^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} $

${\text{PQ}} = \sqrt {9 + 1 + 4} $

${\text{PQ}} = \sqrt {14} $

Now,${\text{QR}} = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} $

${\text{QR}} = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 4} \right)}^2}} $

${\text{QR}} = \sqrt {36 + 4 + 16} $

${\text{QR}} = \sqrt {56} $

By square root,

${\text{QR}} = 2\sqrt {14} $

Now, ${\text{PR}} = \sqrt {{{\left( {7 - \left( { - 2} \right)} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}} $

${\text{PR}} = \sqrt {{{\left( {7 + 2} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}} $

${\text{PR}} = \sqrt {{9^2} + {{\left( { - 3} \right)}^2} + {{\left( { - 6} \right)}^2}} $

${\text{PR}} = \sqrt {81 + 9 + 36} $

${\text{QR}} = \sqrt {126} $

By square root,

${\text{QR}} = 3\sqrt {14} $

So, ${\text{PQ}} + {\text{QR}} = {\text{PR}}$

Thus, the points ${\text{P,Q}}$and ${\text{R}}$ are collinear.

Hence, the points $\left( { - 2,3,5} \right),\left( {1,2,3} \right)$ and $\left( {7,0, - 1} \right)$ are collinear.

3. Verify the following:

(i). $\left( {0,7, - 10} \right),\left( {1,6, - 6} \right)$and$\left( {4,9, - 6} \right)$ are the vertices of an isosceles triangle.

Ans: Let the given points are ${\text{P}}\left( {0,7, - 10} \right),{\text{Q}}\left( {1,6, - 6} \right)$and ${\text{R}}\left( {4,9, - 6} \right)$.

We know that, Aa triangle is called isosceles triangle if two sides of the triangle are equal.

Now, ${\text{PQ}} = \sqrt {{{\left( {1 - 0} \right)}^2} + {{\left( {6 - 7} \right)}^2} + {{\left( { - 6 - \left( { - 10} \right)} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( {1 - 0} \right)}^2} + {{\left( {6 - 7} \right)}^2} + {{\left( { - 6 + 10} \right)}^2}} $

${\text{PQ}} = \sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2}} $

${\text{PQ}} = \sqrt {1 + 1 + 16} $

${\text{PQ}} = \sqrt {18} $

By square root,

${\text{PQ}} = 3\sqrt 2 $

Now,${\text{QR}} = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( { - 6 - \left( { - 6} \right)} \right)}^2}} $

${\text{QR}} = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( { - 6 + 6} \right)}^2}} $

${\text{QR}} = \sqrt {{3^2} + {3^2} + {0^2}} $

${\text{QR}} = \sqrt {9 + 9 + 0} $

${\text{QR}} = \sqrt {18} $

By square root,

${\text{QR}} = 3\sqrt 2 $

Now, ${\text{PR}} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {9 - 7} \right)}^2} + {{\left( { - 6 - \left( { - 10} \right)} \right)}^2}} $

${\text{PR}} = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {9 - 7} \right)}^2} + {{\left( { - 6 + 10} \right)}^2}} $

${\text{PR}} = \sqrt {{4^2} + {2^2} + {4^2}} $

${\text{PR}} = \sqrt {16 + 4 + 16} $

${\text{PR}} = \sqrt {36} $

By square root,

${\text{PR}} = 6$

Thus, ${\text{PQ}} = {\text{QR}}$

Hence, $\left( {0,7, - 10} \right),\left( {1,6, - 6} \right)$ and $\left( {4,9, - 6} \right)$ are the vertices of an isosceles triangle.

(ii). $\left( {0,7,10} \right),\left( { - 1,6,6} \right)$ and $\left( { - 4,9,6} \right)$ are the vertices of a right angled triangle.

Ans: Let the given points are${\text{P}}\left( {0,7,10} \right),{\text{Q}}\left( { - 1,6,6} \right)$and ${\text{R}}\left( { - 4,9,6} \right)$ .

We know that, in a right angled triangle 

${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Base}}} \right)^2} + {\left( {{\text{Altitude}}} \right)^2}$

Now, ${\text{PQ}} = \sqrt {{{\left( { - 1 - 0} \right)}^2} + {{\left( {6 - 7} \right)}^2} + {{\left( {6 - 10} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 4} \right)}^2}} $

${\text{PQ}} = \sqrt {1 + 1 + 16} $

${\text{PQ}} = \sqrt {18} $

By square root,

${\text{PQ}} = 3\sqrt 2 $

Now,${\text{QR}} = \sqrt {{{\left( { - 4 - \left( { - 1} \right)} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( {6 - 6} \right)}^2}} $

${\text{QR}} = \sqrt {{{\left( { - 4 + 1} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( {6 - 6} \right)}^2}} $

${\text{QR}} = \sqrt {{{\left( { - 3} \right)}^2} + {3^2} + {0^2}} $

${\text{QR}} = \sqrt {9 + 9 + 0} $

${\text{QR}} = \sqrt {18} $

By square root,

${\text{QR}} = 3\sqrt 2 $

Now, ${\text{PR}} = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {9 - 7} \right)}^2} + {{\left( {6 - 10} \right)}^2}} $

${\text{PR}} = \sqrt {{{\left( { - 4} \right)}^2} + {2^2} + {{\left( { - 4} \right)}^2}} $

${\text{PR}} = \sqrt {16 + 4 + 16} $

${\text{PR}} = \sqrt {36} $

By square root,

${\text{PR}} = 6$

Now, ${\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2} = {\left( {3\sqrt 2 } \right)^2} + {\left( {3\sqrt 2 } \right)^2}$

${\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2} = 18 + 18$

${\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2} = 36$

Again, ${\text{P}}{{\text{R}}^2} = 36$

Thus, ${\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2} = {\text{P}}{{\text{R}}^2}$

Hence, $\left( {0,7,10} \right),\left( { - 1,6,6} \right)$ and $\left( { - 4,9,6} \right)$are the vertices of a right angled triangle.

(iii). $\left( { - 1,2,1} \right),\left( {1, - 2,5} \right),\left( {4, - 7,8} \right)$and$\left( {2, - 3,4} \right)$ are the vertices of a parallelogram.

Ans: Let the given points are${\text{P}}\left( { - 1,2,1} \right),{\text{Q}}\left( {1, - 2,5} \right),{\text{R}}\left( {4, - 7,8} \right)$and ${\text{S}}\left( {2, - 3,4} \right)$ .

We know that, when the opposite sides of the quadrilateral are equal, is called parallelogram.

Now, ${\text{PQ}} = \sqrt {{{\left( {1 - \left( { - 1} \right)} \right)}^2} + {{\left( { - 2 - 2} \right)}^2} + {{\left( {5 - 1} \right)}^2}} $

${\text{PQ}} = \sqrt {{{\left( {1 + 1} \right)}^2} + {{\left( { - 2 - 2} \right)}^2} + {{\left( {5 - 1} \right)}^2}} $

${\text{PQ}} = \sqrt {{2^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}} $

${\text{PQ}} = \sqrt {4 + 16 + 16} $

${\text{PQ}} = \sqrt {36} $

By square root,

${\text{PQ}} = 6$

Now,${\text{QR}} = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( { - 7 - \left( { - 2} \right)} \right)}^2} + {{\left( {8 - 5} \right)}^2}} $

${\text{QR}} = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( { - 7 + 2} \right)}^2} + {{\left( {8 - 5} \right)}^2}} $

${\text{QR}} = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 5} \right)}^2} + {3^2}} $

${\text{QR}} = \sqrt {9 + 25 + 9} $

${\text{QR}} = \sqrt {53} $

Now, ${\text{RS}} = \sqrt {{{\left( {2 - 4} \right)}^2} + {{\left( { - 3 - \left( { - 7} \right)} \right)}^2} + {{\left( {4 - 8} \right)}^2}} $

${\text{RS}} = \sqrt {{{\left( {2 - 4} \right)}^2} + {{\left( { - 3 + 7} \right)}^2} + {{\left( {4 - 8} \right)}^2}} $

${\text{RS}} = \sqrt {{{\left( { - 2} \right)}^2} + {4^2} + {{\left( { - 4} \right)}^2}} $

${\text{RS}} = \sqrt {4 + 16 + 16} $

${\text{RS}} = \sqrt {36} $

By square root,

${\text{RS}} = 6$

Now, ${\text{SP}} = \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( {2 - \left( { - 3} \right)} \right)}^2} + {{\left( {1 - 4} \right)}^2}} $

${\text{SP}} = \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( {2 + 3} \right)}^2} + {{\left( {1 - 4} \right)}^2}} $

${\text{SP}} = \sqrt {{{\left( { - 3} \right)}^2} + {5^2} + {{\left( { - 3} \right)}^2}} $

${\text{SP}} = \sqrt {9 + 25 + 9} $

${\text{SP}} = \sqrt {53} $

Thus, ${\text{PQ = RS}}$

${\text{QR = SP}}$

Hence, $\left( { - 1,2,1} \right),\left( {1, - 2,5} \right),\left( {4, - 7,8} \right)$ and $\left( {2, - 3,4} \right)$ are the vertices of a parallelogram.

4. Find the equation of the set of points which are equidistant from the points$\left( {1,2,3} \right)$and $\left( {3,2, - 1} \right)$.

Ans: Let the point is ${\text{P}}$ and coordinate of the point${\text{P}}$ be $\left( {x,y,z} \right)$

Also let the given points are $A\left( {1,2,3} \right)$and $B\left( {3,2, - 1} \right)$.

Now, ${\text{PA}} = \sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2} + {{\left( {z - 3} \right)}^2}} $

Also, ${\text{PB}} = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 2} \right)}^2} + {{\left( {z - \left( { - 1} \right)} \right)}^2}} $

${\text{PB}} = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 2} \right)}^2} + {{\left( {z + 1} \right)}^2}} $

Given that, ${\text{PA = PB}}$

So, $\sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 2} \right)}^2} + {{\left( {z - 3} \right)}^2}}  = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 2} \right)}^2} + {{\left( {z + 1} \right)}^2}} $

By squaring,

${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 3} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z + 1} \right)^2}$

Subtracting  ${\left( {y - 2} \right)^2}$ from both sides,

${\left( {x - 1} \right)^2} + {\left( {z - 3} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {z + 1} \right)^2}$

${\left( {x - 1} \right)^2} + {\left( {z - 3} \right)^2} - {\left( {x - 3} \right)^2} - {\left( {z + 1} \right)^2} = 0$

${x^2} - 2x + 1 + {z^2} - 6z + 9 - {x^2} + 6x - 9 - {z^2} - 2z - 1 = 0$

By simplifying,

$4x - 8z = 0$

Hence, the required equation is $4x - 8z = 0$ .

5. Find the equation of the set of points P, the sum of whose distances from${\text{A}}\left( {4,0,0} \right)$and${\text{B}}\left( { - 4,0,0} \right)$ is equal to $10$ .

Ans: Let thecoordinate of the point${\text{P}}$ be $\left( {x,y,z} \right)$

The given points are ${\text{A}}\left( {4,0,0} \right)$and ${\text{B}}\left( { - 4,0,0} \right)$.

Now, ${\text{PA}} = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}} $

${\text{PA}} = \sqrt {{{\left( {x - 4} \right)}^2} + {y^2} + {z^2}} $

Also, ${\text{PB}} = \sqrt {{{\left( {x - \left( { - 4} \right)} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}} $

${\text{PB}} = \sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}} $

According the information,

${\text{PA + PB = }}10$

$\sqrt {{{\left( {x - 4} \right)}^2} + {y^2} + {z^2}}  + \sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  = 10$

$\sqrt {{{\left( {x - 4} \right)}^2} + {y^2} + {z^2}}  = 10 - \sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}} $

By squaring both sides,

${\left[ {\sqrt {{{\left( {x - 4} \right)}^2} + {y^2} + {z^2}} } \right]^2} = {\left[ {10 - \sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}} } \right]^2}$

${\left( {x - 4} \right)^2} + {y^2} + {z^2} = {10^2} - 2 \times 10 \times \sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  + {\left[ {\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}} } \right]^2}$

${\left( {x - 4} \right)^2} + {y^2} + {z^2} = 100 - 20\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  + {\left( {x + 4} \right)^2} + {y^2} + {z^2}$

${x^2} - 8x + 16 + {y^2} + {z^2} = 100 - 20\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  + {x^2} + 8x + 16 + {y^2} + {z^2}$

Simplify,

$ - 8x = 100 - 20\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  + 8x$

$20\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  = 100 + 8x + 8x$

$20\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  = 100 + 16x$

Dividing by $4$ ,

$5\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}}  = 25 + 4x$

By squaring both sides,

${\left[ {5\sqrt {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}} } \right]^2} = {\left( {25 + 4x} \right)^2}$

Since, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

$25\left( {{{\left( {x + 4} \right)}^2} + {y^2} + {z^2}} \right) = 625 + 200x + 16{x^2}$

$25\left( {{x^2} + 8x + 16 + {y^2} + {z^2}} \right) = 625 + 200x + 16{x^2}$

$25{x^2} + 200x + 400 + 25{y^2} + 25{z^2} = 625 + 200x + 16{x^2}$

By simplifying,

$25{x^2} + 200x + 400 + 25{y^2} + 25{z^2} - 625 - 200x - 16{x^2} = 0$

$9{x^2} + 25{y^2} + 25{z^2} - 225 = 0$

Hence, the required equation is $9{x^2} + 25{y^2} + 25{z^2} - 225 = 0$ .

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

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