# Important Questions for Class 11 Chapter 12 - Introduction to Three dimensional Geometry

The Important Questions of class 11 carries different types of questions like word problems, multiple choice questions, etc. in math which makes the preparation all the more comfortable. Students can solve the questions and practice every day.

**Important questions based on NCERT syllabus for Class 11 Chapter 12 - Introduction to Three dimensional Geometry:**

*Question 1*: FInd the direction of the cosines of the line passing through the points A(3, -6, 5) and B(-1, 4, 9)

*Solution*: We know the direction cosines of the line passing through two points A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2}) are given by (x_{2} - x_{1})/AB, (y_{2} - y_{1})/AB, (z_{2} - z_{1})/AB

Where AB = √(x2 - x_{1})^2 + (y2 - y_{1})^2 + (z2 - z_{1})^2

Here A = (3, -6, 5) and B = (-1, 4, 9)

So, AB = √(-1-3)^2 + (4+6)^2 +(9-5)^2

AB = √(-4)^2 + (10)^2 + (4)^2

AB = √16 + 100 + 16

AB = √132 = √4 x 33 = 2√33

Thus, the direction cosines of the line joining two points is

(x_{2} - x_{1})/AB, (y_{2} - y_{1})/AB, (z_{2} - z_{1})/AB

= (-4/2√33 , 10/2√33 , 4/2√33)

= (-2/√33 , 5/√33, 2/√33)

*Question 2*: Find the points of trisection of the points P(4, 2, -6) and Q (10, -16, 6).

*Solution*: Let A and B be the points that trisect the line segment joining points P(4, 2, -6) and Q(10, -16, 6).

Point A divides PQ in the ratio 1:2, therefore by section formula the corrdinates of point A are given by

(1(10)+2(4)/1+2 , 1(-16) + 2(2)/ 1+2 , 1(6) + 2(-6)/1+2) = (18/3, -12/3, -6/3) = (6,-4,-2)

Point B divides PQ in the ratio 2:1, therefore by section formula the corrdinates of point B are given by

(2(10)+1(4)/2+1 , 2(-16) + 1(2)/ 2+1 , 2(6) + 1(-6)/2+1) = (24/3, -30/3, -6/3) = (8, -10, -2)

Thus (6,-4,-2) and (8, -10, -2) are the points of trisection of the points P(4, 2, -6) and Q (10, -16, 6).

*Question 3*: The centroid of a triangle ABC is (1, 1, 1). Given the vertices A(3,-5,7) and B(-1,7,6) then find the vertex C.

*Solution*: Given ABC is a triangle with the vertices A(3,-5,7) and B(-1,7,6)

Let G be the centroid of the triangle ABC. So G(1, 1, 1)

We know that co-ordinate of centroid whose vertices are (x1, y_{1}, z_{1}),(x_{2}, y_{2}, z_{2}) (x3, y_{3}, z_{3}) is

(x_{1}, y_{1}, z_{1})/ 3 (x_{2}, y_{2}, z_{2})/3 (x3, y3, z3)/3

Therefore, the coordinates of G

(1, 1, 1) = ((3+(-1)+x)/3 , (-5+7+y)/3 , (7+6+z)/3)

= ((2+x)/3 , (2+y)/3, (13+z)/3)

for x co-ordinate 1 = 2+x/3

=> 3 = 2+x

=> 3-2 = x

=> x = 1

for y co-ordinate 1 = (2+y)/3

=> 3 = 2+y

=> 3-2 = y

=> y = 1

for z co-ordinate 1 = (13+z)/3

=> 3 = 13 + z

=> 3 - 13 = z

=> z = -10

Hence the required coordinate of C is (1, 1, -10)