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Vidaara.orgClass 12 · Mathematics
CodeVID-M12-11-CT
Three-Dimensional Geometry — Full Chapter Test
Chapter: Three-Dimensional Geometry
Topic: Complete chapter — all topics
Maximum Marks: 35
Time: 80 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
  • Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
  • Show all working for Sections B, C and D. Final answers are given in the Answer Key.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
If a line makes angles $\alpha,\beta,\gamma$ with the coordinate axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ equals:
  • A.$1$
  • B.$2$
  • C.$3$
  • D.$0$
2.
The distance between the parallel planes $2x+3y-6z=14$ and $2x+3y-6z=-7$ is:
  • A.$1$
  • B.$3$
  • C.$7$
  • D.$21$
3.
Assertion (A): The planes $2x-y+z=3$ and $x+y-z=5$ are perpendicular.
Reason (R): Two planes $a_1x+b_1y+c_1z=d_1$ and $a_2x+b_2y+c_2z=d_2$ are perpendicular if and only if $a_1a_2+b_1b_2+c_1c_2=0$.
  • A.Both A and R are true and R is the correct explanation of A
  • B.Both A and R are true but R is not the correct explanation of A
  • C.A is true but R is false
  • D.A is false but R is true
4.
The angle $\theta$ between the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{6}$ and the plane $10x+2y-11z=3$ satisfies:
  • A.$\sin\theta=\dfrac{21}{8}$
  • B.$\cos\theta=\dfrac{8}{21}$
  • C.$\sin\theta=\dfrac{8}{21}$
  • D.$\tan\theta=\dfrac{8}{21}$
5.
The distance of the point $(3,4,5)$ from the $y$-axis is:
  • A.$\sqrt{41}$
  • B.$\sqrt{34}$
  • C.$5$
  • D.$4$
Section B — Short Answer I 4 × 2 = 8 marks
6.
Find the direction cosines of the line passing through the points $P(-2,4,-5)$ and $Q(1,2,3)$.
7.
Find the equation of the plane through the point $(1,2,3)$ and parallel to the plane $2x-3y+z=5$.
8.
Find the angle between the lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{2}$ and $\dfrac{x+1}{2}=\dfrac{y}{3}=\dfrac{z-5}{6}$.
9.
Find the intercepts made by the plane $2x+y-z=5$ on the coordinate axes.
Section C — Short Answer II 4 × 3 = 12 marks
10.
Find the shortest distance between the lines $\vec r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)$ and $\vec r=(2\hat i-\hat j-\hat k)+\mu(2\hat i+\hat j+2\hat k)$.
11.
Find the equation of the plane through the point $(1,0,-2)$ and perpendicular to each of the planes $2x+y-z=2$ and $x-y-z=3$.
12.
Show that the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-4}{5}=\dfrac{y-1}{2}=z$ intersect, and find their point of intersection.
13.
Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\vec r=2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)$ and the plane $x-y+z=5$.
Section D — Long Answer / Case Based 2 × 5 = 10 marks
14.
Find the image of the point $(1,6,3)$ in the line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$.
15.
Case study. A ramp lies in the plane $2x-y+2z=9$. A sensor is placed at the point $P(1,2,3)$.
(a) Find the perpendicular distance of $P$ from the ramp. [2]
(b) Find the coordinates of the foot of the perpendicular from $P$ to the plane. [2]
(c) Find the image of $P$ in the plane. [1]

Answer Key

Section A — Multiple Choice Questions
  1. (B) $2$
  2. (B) $3$
  3. (A) Both A and R are true and R is the correct explanation of A
  4. (C) $\sin\theta=\dfrac{8}{21}$
  5. (B) $\sqrt{34}$
Section B — Short Answer I
  1. Direction ratios: $(1+2,\,2-4,\,3+5)=(3,-2,8)$, with $\sqrt{9+4+64}=\sqrt{77}$. Direction cosines: $\dfrac{3}{\sqrt{77}},\ \dfrac{-2}{\sqrt{77}},\ \dfrac{8}{\sqrt{77}}$.
  2. A parallel plane has the same normal: $2(x-1)-3(y-2)+(z-3)=0$, i.e. $2x-3y+z+1=0$.
  3. With direction ratios $(1,2,2)$ and $(2,3,6)$: $\cos\theta=\dfrac{2+6+12}{3\cdot7}=\dfrac{20}{21}$. So $\theta=\cos^{-1}\dfrac{20}{21}$.
  4. Writing the plane as $\dfrac{x}{5/2}+\dfrac{y}{5}+\dfrac{z}{-5}=1$, the intercepts are $\dfrac52$ on the $x$-axis, $5$ on the $y$-axis and $-5$ on the $z$-axis.
Section C — Short Answer II
  1. $\vec a_2-\vec a_1=\hat i-3\hat j-2\hat k$. $\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix}=-3\hat i+3\hat k$, with magnitude $3\sqrt2$. $(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=-3-6=-9$. Shortest distance $=\dfrac{9}{3\sqrt2}=\dfrac{3\sqrt2}{2}$ units.
  2. The normal is $\vec n_1\times\vec n_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&1&-1\\1&-1&-1\end{vmatrix}=-2\hat i+\hat j-3\hat k$. The plane is $-2(x-1)+y-3(z+2)=0$, i.e. $2x-y+3z+4=0$.
  3. Points on the lines: $(2\lambda+1,\,3\lambda+2,\,4\lambda+3)$ and $(5\mu+4,\,2\mu+1,\,\mu)$. From the $z$-coordinates, $\mu=4\lambda+3$. From the $x$-coordinates, $2\lambda+1=5\mu+4\Rightarrow2\lambda=20\lambda+18\Rightarrow\lambda=-1$, so $\mu=-1$. The $y$-coordinates: $3(-1)+2=-1$ and $2(-1)+1=-1$ agree, so the lines intersect at $(-1,-1,-1)$.
  4. A point on the line is $(2+3\lambda,\,-1+4\lambda,\,2+2\lambda)$. Substituting in the plane: $(2+3\lambda)-(-1+4\lambda)+(2+2\lambda)=5\Rightarrow5+\lambda=5\Rightarrow\lambda=0$. The point is $(2,-1,2)$. Distance from $(-1,-5,-10)$: $\sqrt{9+16+144}=13$ units.
Section D — Long Answer / Case Based
  1. A point on the line is $Q(\lambda,\,2\lambda+1,\,3\lambda+2)$. For $Q$ to be the foot of the perpendicular from $P(1,6,3)$, $\vec{PQ}=(\lambda-1,\,2\lambda-5,\,3\lambda-1)$ must satisfy $\vec{PQ}\cdot(1,2,3)=0$: $\lambda-1+4\lambda-10+9\lambda-3=0\Rightarrow\lambda=1$. The foot is $Q(1,3,5)$. Since $Q$ is the midpoint of $PP'$, $P'=2Q-P=(1,0,7)$.
  2. (a) $d=\dfrac{|2(1)-2+2(3)-9|}{\sqrt{4+1+4}}=\dfrac{3}{3}=1$ unit. (b) The line through $P$ along the normal is $(1+2k,\,2-k,\,3+2k)$. Putting it in the plane: $2(1+2k)-(2-k)+2(3+2k)=9\Rightarrow9k+6=9\Rightarrow k=\dfrac13$. The foot is $\left(\dfrac53,\dfrac53,\dfrac{11}{3}\right)$. (c) The image is at $k=\dfrac23$: $\left(\dfrac73,\dfrac43,\dfrac{13}{3}\right)$.
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