NCERT Solutions for Class 11 Maths Chapter 12: Introduction to Three Dimensional Geometry - 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

Opting for the NCERT solutions for Ex 12.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 12 Exercise 12.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 12 Exercise 12.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 12 Exercise 12.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.