Vidaara.orgClass 12 · Mathematics
CodeVID-M12-11-CT
Three-Dimensional Geometry — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- 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.
Direction cosines satisfy:
- A.$l+m+n=1$
- B.$l^2+m^2+n^2=1$
- C.$lmn=1$
- D.$l^2+m^2+n^2=0$
2.
The vector equation of a line is:
- A.$\vec r=\vec a+\lambda\vec b$
- B.$\vec r\cdot\vec n=d$
- C.$\vec r=\vec a\times\vec b$
- D.$|\vec r|=1$
3.
Two lines are perpendicular when their direction vectors satisfy:
- A.$\vec b_1\times\vec b_2=\vec0$
- B.$\vec b_1\cdot\vec b_2=0$
- C.$\vec b_1=\vec b_2$
- D.$|\vec b_1|=|\vec b_2|$
4.
Direction cosines of a line with DRs $1,2,2$ are:
- A.$1,2,2$
- B.$\tfrac13,\tfrac23,\tfrac23$
- C.$\tfrac12,\tfrac12,\tfrac12$
- D.$\tfrac15,\tfrac25,\tfrac25$
5.
In $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z}{4}$, the direction ratios are:
- A.$(1,-3,0)$
- B.$(2,-1,4)$
- C.$(1,3,0)$
- D.$(2,1,4)$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Find the direction cosines of a line with DRs $2,3,6$.
7.
Write the vector equation of the line through $(1,0,2)$ with direction $\hat i+\hat j-\hat k$.
8.
Find the angle between lines with directions $\hat i+\hat j+\hat k$ and $\hat i-\hat j+\hat k$.
9.
Find the direction ratios of the line joining $(2,3,4)$ and $(5,6,7)$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Find the direction cosines of the line joining $(1,0,0)$ and $(0,1,1)$.
11.
Find the vector equation of the line through $(2,-1,1)$ and $(4,1,3)$.
12.
Find the angle between lines with directions $(1,2,2)$ and $(2,2,1)$.
13.
A line makes equal angles with the three axes. Find that angle.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Show that the points $A(2,3,-4),\ B(1,-2,3),\ C(3,8,-11)$ are collinear.
15.
Find the vector and Cartesian equations of the line through $(1,2,3)$ parallel to $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}$.
Answer Key
Section A — Multiple Choice Questions
- (B) $l^2+m^2+n^2=1$
- (A) $\vec r=\vec a+\lambda\vec b$
- (B) $\vec b_1\cdot\vec b_2=0$
- (B) $\tfrac13,\tfrac23,\tfrac23$
- (B) $(2,-1,4)$
Section B — Short Answer (2 marks)
- $\tfrac27,\tfrac37,\tfrac67$.
- $\vec r=(\hat i+2\hat k)+\lambda(\hat i+\hat j-\hat k)$.
- $\cos^{-1}\!\left(\tfrac13\right)$.
- $(3,3,3)$ or $(1,1,1)$.
Section C — Short Answer (3 marks)
- $\left(-\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3}\right)$.
- $\vec r=(2\hat i-\hat j+\hat k)+\lambda(\hat i+\hat j+\hat k)$.
- $\cos^{-1}\!\left(\tfrac89\right)$.
- $\cos^{-1}\!\left(\tfrac{1}{\sqrt3}\right)$.
Section D — Long Answer (5 marks)
- Collinear.
- $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+6\hat k)$; $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{6}$.
Generated by Vidaara.org · Assignment VID-M12-11-CT · vidaara.org