# Class 11 Maths MCQ – Three Dimensional Geometry – Section Formula

This set of Class 11 Maths Chapter 12 Multiple Choice Questions & Answers (MCQs) focuses on “Three Dimensional Geometry – Section Formula”.

1. Find midpoint of (1, 4, 6) and (5, 8, 10).
a) (6, 12, 8)
b) (3, 6, 8)
c) (1, 9, 12)
d) (4, 9, 12)

Explanation: We know, midpoint of (x1, y1, z1) and (x2, y2, z2) is (x1+x2) /2, (y1+y2) /2, (z1+z2)/2).
So, midpoint of (1, 4, 6) and (5, 8, 10) is ((1+5)/ 2, (4+8)/ 2, (6+10)/2) is (3, 6, 8).

2. The coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio 2:1 is ____________________
a) (3, 4, 5)
b) (5, 4, 3)
c) (5, 3, 4)
d) (4, 5, 3)

Explanation: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is $$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$$.
So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio 2:1 is $$(\frac{2*4+1*1}{2+1},\frac{2*5+1*2}{2+1},\frac{2*6+1*3}{2+1})$$ = (3, 4, 5).

3. In which ratio (3, 4, 5) divides the line segment joining (1, 2, 3) and (4, 5, 6) internally?
a) 1:2
b) 2:1
c) 3:4
d) 4:3

Explanation: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m: n is $$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$$.
Let the ratio be k : 1.So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio k: 1 is $$(\frac{k*4+1*1}{k+1},\frac{k*5+1*2}{k+1},\frac{k*6+1*3}{k+1})$$
=> $$(\frac{k*4+1*1}{k+1},\frac{k*5+1*2}{k+1},\frac{k*6+1*3}{k+1})$$ is same as (3, 4, 5).
=> (4k+1)/(k+1) = 3
=> 4k+1 = 3k+3
=> k = 2
So, ratio is 2:1.

4. The coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) externally in the ratio 2:1 is ____________________
a) (4, 5, 6)
b) (6, 8, 9)
c) (7, 8, 9)
d) (8, 6, 4)

Explanation: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) externally in the ratio m : n is $$(\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n},\frac{mz_2-nz_1}{m-n})$$.
So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) externally in the ratio 2:1 is $$(\frac{2*4-1*1}{2-1},\frac{2*5-1*2}{2-1},\frac{2*6-1*3}{2-1})$$ = (7, 8, 9).

5. If coordinates of vertices of a triangle are (7, 6, 4), (5, 4, 6), (9, 5, 8), find the coordinates of centroid of the triangle.
a) (7, 5, 3)
b) (7, 3, 5)
c) (5, 3, 7)
d) (3, 5, 7)

Explanation: If coordinates of vertices of a triangle are (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) the coordinates of centroid of the triangle are ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3)
So, coordinates of centroid of the given triangle are ((7+5+9)/3, (6+4+5)/3, (4+6+8)/3) = (7, 5, 3).
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6. The ratio in which line joining (1, 2, 3) and (4, 5, 6) divide X-Y plane is ________
a) 2
b) -2
c) 1/2
d) -1/2

Explanation: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is $$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$$.
Let ratio be k : 1.
So, z-coordinate of the point will be (k*6+1*3)/(k+1).
We know, for X-Y plane, z coordinate is zero.
(6k+1*3)/(k+1) = 0 => k=-1/2

7. Find the points which trisects the line joining (4, 9, 8) and (13, 27, -4).
a) (7, 4, 15)
b) (7, 15, 4)
c) (4, 15, 7)
d) (4, 7, 15)

Explanation: Points which trisect the line divides it into 2:1 and 1:2.
The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is $$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$$.
For 1:2, coordinates of point are $$(\frac{1*13+2*4}{1+2},\frac{1*27+2*9}{1+2},\frac{-4+2*8}{1+2})$$ = (7, 15, 4)
For 2:1, coordinates of point are $$(\frac{2*13+1*4}{1+2},\frac{2*27+1*9}{1+2},\frac{-8+1*8}{1+2})$$ = (10, 21, 0)

8. Find the points which trisects the line joining (4, 9, 8) and (13, 27, -4).
a) (0, 21, 10)
b) (0, 21, 4)
c) (10, 21, 0)
d) (4, 4, 0)

Explanation: Points which trisect the line divides it into 2:1 and 1:2.
The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is $$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$$.
For 1:2, coordinates of point are $$(\frac{1*13+2*4}{1+2},\frac{1*27+2*9}{1+2},\frac{-4+2*8}{1+2})$$ = (7, 15, 4)
For 2:1, coordinates of point are $$(\frac{2*13+1*4}{1+2},\frac{2*27+1*9}{1+2},\frac{-8+1*8}{1+2})$$ = (10, 21, 0)

9. If P (2, 3, 9), Q (2, 5, 5) and R (8, 5, 3) are vertices of a triangle then find the length of median through P.
a) $$\sqrt{24}$$
b) $$\sqrt{38}$$
c) $$\sqrt{11}$$
d) $$\sqrt{53}$$

Explanation: We know, midpoint of (x1, y1, z1) and (x2, y2, z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Midpoint of line QR is (5, 5, 4).
Length of median through P is distance between midpoint of QR and P i.e. $$\sqrt{(5-2)^2+(5-3)^2+(4-9)^2} = \sqrt{(3)^2+(2)^2+(-5)^2} = \sqrt{38}$$

10. If P (2, 3, 9), Q (2, 5, 5) and R (8, 5, 3) are vertices of a triangle then find the length of median through Q.
a) $$\sqrt{24}$$
b) $$\sqrt{38}$$
c) $$\sqrt{11}$$
d) $$\sqrt{53}$$

Explanation: We know, midpoint of (x1, y1, z1) and (x2, y2, z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Midpoint of line PR is (5, 4, 6).
Length of median through Q is distance between midpoint of PR and Q i.e. $$\sqrt{(5-2)^2+(4-5)^2+(6-5)^2} = \sqrt{(3)^2+(-1)^2+(1)^2} = \sqrt{11}$$.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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