Vidaara.orgClass 11 · Mathematics
CodeVID-M11-11-SEC-01
Section Formula in 3-D — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The midpoint is the ___ of the coordinates.
- A.sum
- B.average
- C.product
- D.difference
2.
The midpoint of $(2,0,0)$ and $(0,2,0)$ is:
- A.$(1,1,0)$
- B.$(2,2,0)$
- C.$(0,0,0)$
- D.$(1,0,1)$
3.
The internal section formula for $x$ (ratio $m:n$) is:
- A.$\tfrac{mx_2+nx_1}{m+n}$
- B.$\tfrac{x_1+x_2}{2}$
- C.$mx_1+nx_2$
- D.$\tfrac{mx_1-nx_2}{m-n}$
4.
The centroid of a triangle is the ___ of the vertices.
- A.sum
- B.average
- C.midpoint of two
- D.product
5.
The midpoint of $(1,2,3)$ and $(3,4,5)$ is:
- A.$(2,3,4)$
- B.$(4,6,8)$
- C.$(1,1,1)$
- D.$(2,2,2)$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Find the midpoint of $(1,2,3)$ and $(5,6,7)$.
7.
Find the point dividing $(1,2,3)$ and $(4,5,6)$ in the ratio $1:2$.
8.
Find the centroid of the triangle $(0,0,0),(3,0,0),(0,3,0)$.
9.
Find the midpoint of $(2,-4,6)$ and $(-2,4,-6)$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Find the point dividing $(2,3,4)$ and $(6,7,8)$ in the ratio $3:1$.
11.
In what ratio does $(2,4,6)$ divide the segment from $(1,2,3)$ to $(4,8,12)$?
12.
Find the centroid of the triangle with vertices $(1,2,3),(4,5,6),(7,8,9)$.
13.
Find the midpoint of the segment joining $(a,b,c)$ and $(-a,-b,-c)$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Find the ratio in which $(2,4,6)$ divides the segment joining $(1,2,3)$ and $(4,8,12)$, and verify your answer.
15.
The vertices of a triangle are $A(3,-1,2),\ B(1,-1,-3),\ C(4,-3,1)$. Find the centroid.
Answer Key
Section A — Multiple Choice Questions
- (B) average
- (A) $(1,1,0)$
- (A) $\tfrac{mx_2+nx_1}{m+n}$
- (B) average
- (A) $(2,3,4)$
Section B — Short Answer (2 marks)
- $(3,4,5)$.
- $(2,3,4)$.
- $(1,1,0)$.
- $(0,0,0)$.
Section C — Short Answer (3 marks)
- $(5,6,7)$.
- $1:2$.
- $(4,5,6)$.
- $(0,0,0)$.
Section D — Long Answer (5 marks)
- $1:2$ (verified by the section formula).
- $\left(\tfrac{8}{3},-\tfrac{5}{3},0\right)$.
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