Sample QuestionsThree Dimensional Geometry questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z = 0 is:
- A
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}+\lambda(\hat{\text{i}}+\hat{\text{j}})$
- ✓
$\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
- C
$\vec{\text{r}}=\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}+\lambda\hat{\text{k}}$
- D
$\vec{\text{r}}=\lambda\hat{\text{k}}$
Answer: B.
View full solution →The cosines of the angle between any two diagonals of a cube is:
- ✓
$\frac{1}{3}$
- B
$\frac{1}{2}$
- C
$\frac{2}{3}$
- D
$\frac{1}{\sqrt{3}}$
Answer: A.
View full solution →The xy-plane divided the line joining the point (-1, 3, 4) and (2, -5, 6)
- A
Internally in the ratio 2 : 3
- ✓
Externally in the ratio 2 : 3
- C
Internally in the ratio 3 : 2
- D
Externally in the ratio 3 : 2
Answer: B.
View full solution →A line OP where O = (0, 0, 0) makes equal angles with ox, oy, oz. The point on OP, which is at a distance of 6 units from O is:
- ✓
$\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
- B
$\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
- C
$-\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
- D
$-\big(6\sqrt{3},-6\sqrt{3},6\sqrt{3}\big)$
Answer: A.
View full solution →Choose the correct answer from the given four options.
The locus represented by xy + yz = 0 is:
- A
A pair of perpendicular lines.
- B
A pair of parallel lines.
- C
A pair of parallel planes.
- ✓
A pair of perpendicular planes.
Answer: D.
View full solution →Directions : In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose the correct answer out of the following choices :
Assertion : If the cartesian equation of a line is $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2},$ then its vector form is $\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}).$
Reason : The cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}$ is $\frac{\text{x}+3}{-2}=\frac{\text{y}-4}{4}=\frac{\text{z}+8}{-5}.$
- A
Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- B
Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- ✓
Assertion is correct statement but Reason is wrong statement.
- D
Assertion is wrong statement but Reason is correct statement.
Answer: C.
View full solution →Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : Points $A(4, 0, 4), B(1, 2, 3), C(-2, 4, 2)$ are collinear.
Reason : Three points $\text{A, B, C}$ are collinear if $\text{AB + BC = AC}$ and $\text{AB, BC < AC}$.
Answer: A.
View full solution →Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : The points $(1, 2, 3), (-2, 3, 4)$ and $(7, 0, 1)$ are collinear
Reason : If a line makes angles $\frac{\pi}{2}, \frac{3\pi}{4}$ and $\frac{\pi}{4}$ with $X, Y,$ and $Z -$ axes respectively, then its direction cosines are $0,\frac{-1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$
- A
Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- ✓
Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- C
Assertion is correct statement but Reason is wrong statement.
- D
Assertion is wrong statement but Reason is correct statement.
Answer: B.
View full solution →Find the angle between the lines $\text{2x = 3y = – z and 6x = – y = – 4z.}$
View full solution →If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
$\text{Find} \lambda, \text{if the vectors} \overrightarrow{\text{a}} = \hat{\text{i}} + 3\hat{\text{j}} + \hat{\text{k}}, \overrightarrow{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} - \hat{\text{k}}$ $\text{and} \overrightarrow{\text{c}} = \lambda \hat{\text{j}} + 3\hat{\text{k}} $ $\text{are coplanar}.$
View full solution →Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
View full solution →Write the vector equation of a line given by $\frac{\text{x - 5}}{3}=\frac{\text{y + 4}}{7}=\frac{\text{z - 6}}{2}.$
View full solution →Write the equation of the plane parallel to XOY- plane and passing through the point (2, -3, 5).
Find the vector equation one of following plane.
x + y - z = 5
Find the vector equation one of following plane.
2x - y + 2z = 8
Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
View full solution →Find the equation of the plane passing through the following point:
(2, 3, 4), (-3, 5, 1) and (4, -1, 2)
Find the vector and cartesian equations of the planes:
That passes through the point (1, 4, 6) and the normal vector to the plane is $\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}.$
View full solution →Find the direction cosines of the line passing through two points $(-2, 4, -5)$ and $(1, 2, 3)$.
View full solution →Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=6$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})=9$
View full solution →If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3, -6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}.$
View full solution →Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).
Find the perpendicular distence of the point (1, 0, 0) from the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}.$ Also, find the coordinates of the perpendicular and the equation of the perpendicular.
View full solution →Show that the plane whose vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the line whose vector equation is $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}).$
View full solution →Find the angle between the pairs of lines with direction ratios proportional to $1, 2, -2$ and $-2, 2, 1$
View full solution →Find the equation of the plane through the points $(2, 1, 0), (3, -2, -2)$ and $(3, 1, 7)$.
View full solution →Suppose the floor of a hotel is made up of mirror polished Kota stone. Also, there is a large crystal chandelier attached at the ceiling of the hotel. Consider the floor of the hotel as a plane having equation $x - 2y + 2z = 3$ and crystal chandelier at the point $(3, -2, 1).$
Based on the above information, answer the following questions.
- The d.r'.s of the perpendicular from the point $(3, -2, 1)$ to the plane $x - 2y + 2z = 3$, is:
- $< 1, 2, 2 >$
- $< 1, -2, 2 >$
- $< 2, 1, 2 >$
- $< 2, -1, 2 >$
- The length of the perpendicular from the point $(3, -2, 1)$ to the plane $x - 2y + 2z = 3,$ is:
- $\frac{2}{3}\text{units}$
- 3 units
- 2 units
- None of these
- The equation of the perpendicular from the point $(3, -2, 1)$ to the plane $x - 2y + 2z = 3,$ is:
- $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-1}{2}$
- $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
- $\frac{\text{x}+3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
- None of these
- The equation of plane parallel to the plane $x - 2y + 2z = 3,$ which is at a unit distance from the point $(3, -2, 1)$ is:
- $x - 2y + 2z = 0$
- $x - 2y + 2z = 6$
- $x - 2y + 2z = 12$
- Both $(b)$ and $(c)$
- The image of the point $(3, -2, 1)$ in the given plane is:
- $\Big(\frac{5}{3},\frac{2}{3},\frac{-5}{3}\Big)$
- $\Big(\frac{-5}{3},\frac{-2}{3},\frac{5}{3}\Big)$
- $\Big(\frac{-5}{3},\frac{2}{3},\frac{5}{3}\Big)$
- None of these
View full solution →Consider the following diagram, where the forces in the cable are given. 
Based on the above information, answer the following questions.
- The equation of line along the cable AD is:
- $\frac{\text{x}}{5}=\frac{\text{y}}{4}=\frac{\text{z}-30}{15}$
- $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{\text{z}-30}{15}$
- $\frac{\text{x}}{5}=\frac{\text{y}}{4}=\frac{30-\text{z}}{15}$
- $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
- The length of cable DC is:
- $4\sqrt{61}\text{m}$
- $5\sqrt{61}\text{m}$
- $6\sqrt{61}\text{m}$
- $7\sqrt{61}\text{m}$
- The vector DB is:
- $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
- $6\hat{\text{i}}-4\hat{\text{j}}+30\hat{\text{k}}$
- $6\hat{\text{i}}+4\hat{\text{j}}+30\hat{\text{k}}$
- None of these
- The sum of vectors along the cables is:
- $17\hat{\text{i}}+6\hat{\text{j}}+90\hat{\text{k}}$
- $17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
- $17\hat{\text{i}}+6\hat{\text{j}}-90\hat{\text{k}}$
- None of these
- The sum of distances of points A, B and C from the origin, i.e., OA + OB + OC is:
- $\sqrt{164}+\sqrt{52}+\sqrt{625}$
- $\sqrt{52}+\sqrt{625}+\sqrt{48}$
- $\sqrt{164}+\sqrt{625}+\sqrt{49}$
- None of these
View full solution →Consider the following diagram, where the forces in the cable are given. 
Based on the above information, answer the following questions.
- The cartesian equation of line along EA is:
- $\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}}{12}$
- $\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}-24}{12}$
- $\frac{\text{x}}{-3}=\frac{\text{y}}{3}=\frac{\text{z}-12}{12}$
- $\frac{\text{x}}{3}=\frac{\text{y}}{4}=\frac{\text{z}-24}{12}$
- The vector $\overline{\text{ED}}$ is:
- $8\hat{\text{i}}-6\hat{\text{j}}+24\hat{\text{k}}$
- $-8\hat{\text{i}}-6\hat{\text{j}}+24\hat{\text{k}}$
- $-8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}}$
- $8\hat{\text{i}}+6\hat{\text{j}}+24\hat{\text{k}}$
- The length of the cable EB is:
- 24 units
- 26 units
- 27 units
- 25 units
- The length of cable EC is equal to the length of:
- EA
- EB
- ED
- All of these
- The sum of all vectors along the cables is:
- $96\hat{\text{i}}$
- $96\hat{\text{j}}$
- $-96\hat{\text{k}}$
- $96\hat{\text{k}}$
View full solution →If $a_{1,} b_{1,} c_{1,}$ and $a_{2,}b_{2,} c_2$ are direction ratios of two lines say $L_1$ and $L_2$ respectively. Then $L_1 || L_2$ iff $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ and $\text{L}_1\perp\text{L}_2$ iff $a_1a_2 + b_1b_2 + c_1c_2 = 0.$
Based on the above information, answer the following questions.
- If $l_{1,} m_1, n_{1,}$ and $l_2, m_2, n_2$ are the direction cosines of $L_1$ and $L_2$ respectively, then $L_1$ will be perpendicular to $L_2,$ iff:
- $l_1l_2 + m_1m_2 + n_1n_2 = 0$
- $l_1m_2 + m_1l_2 + n_1n_2 = 0$
- $\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$
- None of these
- If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of $L_1$ and $L_2$ respectively, then $L_1$ will be parallel to $L_2,$ iff:
- $l_1l_2 + m_1m_2 + n_1n_2 = 0$
- $l_1m_2 + m_1l_2 + n_1n_2 = 0$
- $\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$
- $m_1n_2 + m_2n_2 + l_1l_2 = 0$
- The coordinates of the foot of the perpendicular drawn from the point $A(1, 2, 1)$ to the line joining $B(1, 4, 6)$ and $C(5, 4, 4),$ are:
- $(1, 2, 1)$
- $(2, 4, 5)$
- $(3, 4, 5)$
- $(3, 4, 5)$
- The direction ratios of the line which is perpendicular to the lines with direction ratios proportional to $(1, -2, -2)$ and $(0, 2, 1)$ are:
- $< 1, 2, 1 >$
- $< 2,-1, 2 >$
- $< -1,2, 2 >$
- None of these
- The lines $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{-2}=\frac{\text{z}-2}{0}$ and $\frac{\text{x}-1}{1}=\frac{\frac{\text{y}+3}{2}}{\frac{3}{2}}=\frac{\text{z}+5}{2}$ are:
- Parallel.
- Perpendicular.
- Skew lines.
- Non-intersecting.
View full solution →Two motorcycles A and Bare running at the speed more than allowed speed on the road along the lines $\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$ and $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}+\mu(2\hat{\text{i}+\hat{\text{j}}+\hat{\text{k}}}),$ respectively. 
Based on the above information, answer the following questions.
- The cartesian equation of the line along which motorcycle A is running is:
- $\frac{\text{x}+1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{-1}$
- $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{-1}$
- $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{1}$
- None of these
- The direction cosines of line along which motorcycle A is running, are:
- < 1, -2, 1 >
- < I, 2, -1 >
- $<\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}>$
- $<\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$
- The direction ratios of line along which motorcycle Bis running, are:
- < 1, 0, 2 >
- < 2, 1, 0 >
- < 1, 1, 2 >
- < 2, 1, 1 >
- The shortest distance between the gives lines is:
- 4 units
- $2\sqrt{3}\text{ units}$
- $3\sqrt{2}\text{ units}$
- 0 units
- The motorcycles will meet with an accident at the point:
- (-1, 1, 2)
- (2, 1, -1)
- (1, 2, -1)
- Does not exist
View full solution →Fill in the blanks.
The vector equation of the line through the points (3, 4, -7) and (1, -1, 6) is __________.
Fill in the blanks.
The cartesian equation of the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=2$ is ________.
View full solution →Fill in the blanks.
A plane passes through the points (2, 0, 0) (0, 3, 0) and (0, 0, 4). The equation of plane is __________.
The position vectors of two points A and B are $\overrightarrow{\text{OA}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and $\overrightarrow{\text{OB}}=2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}},$ respectively. The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is ___________.
View full solution →The position vectors of two points A and B are $\overrightarrow{\text{OA}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and $\overrightarrow{\text{OB}}=2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}},$ respectively. The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is ___________.
View full solution →State True or False for the following:
The line $\vec{\text{r}}=2\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}+\lambda({\text{i}}-{\text{j}}+2{\text{k}})$ lies in the plane $\vec{\text{r}}\cdot(3\hat{\text{i}}+{\text{j}}-{\text{k}})+2=0$
View full solution →State True or False for the following: The angle between the line $\vec{\text{r}}=(5\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})+\lambda(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})$ and the plane $\vec{\text{r}}(3\hat{\text{i}}-4\hat{\text{j}}-\hat{\text{k}})+5=0$ is $\sin^{-1}\Big(\frac{5}{2\sqrt{91}}\Big).$
View full solution →State True or False for the following:
The equation of a line, which is parallel to $2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and which passes through the point (5, -2, 4) is $\frac{\text{x}-5}{2}=\frac{\text{y}-5}{-1}=\frac{\text{z}-4}{3}.$
View full solution →State True or False for the following:
The intercepts made by the plane 2x - 3y + 5z + 4 = 0 on the co-ordinate axis are $-2, \frac{4}{3},-\frac{4}{5}.$
View full solution →State True or False for the following:
The vector equation of the line $\frac{\text{x}-5}{3}=\frac{\text{y}-4}{7}=\frac{\text{z}-6}{2}$ is $\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda(3\hat{\text{i}}-7\hat{\text{j}}+2\hat{\text{k}}).$
View full solution →