Please refer to below Exam Question for Class 12 Mathematics Chapter 11 Three Dimensional Geometry. These questions and answers have been prepared by expert Class 12 Mathematics teachers based on the latest NCERT Book for Class 12 Mathematics and examination guidelines issued by CBSE, NCERT, and KVS. We have provided Class 12 Mathematics exam questions for all chapters in your textbooks. You will be able to easily learn problems and solutions which are expected to come in the upcoming class tests and exams for standard 10th.

Chapter 11 Three Dimensional Geometry Class 12 Mathematics Exam Question

All questions and answers provided below for Exam Question Class 12 Mathematics Chapter 11 Three Dimensional Geometry are very important and should be revised daily.

Exam Question Class 12 Mathematics Chapter 11 Three Dimensional Geometry

Very Short Answer Type Questions

Question. Write the distance of a point P(a, b, c) from x-axis.
Answer. We have eq. of x-axis as y = 0, z = 0
∴ Distance of P(a, b, c) from x-axis

Question. Find the distance of the point (2, 3, 4) from the x-axis.
Answer. Let B(2, 3, 4) be the given point and A be a point on x-axis s.t. AB is 􀁁 to x-axis.
∴ A ≡(2, 0, 0)
Distance of the point B from x-axis is

Question. If the cartesian equation of a line is

Answer. The cartesian equation of a line is

Question. Find the direction cosines of the line

Answer.

Question. Find the angle between the lines
r̅ = 2î − 5ĵ + k̂ + λ(3î + 2ĵ + 6k̂) and
r̅ = 7î − 6k̂ + μ(î + 2ĵ + 2k̂).
Answer. For r̅ = 2î − 5ĵ + k̂ + λ(3î + 2ĵ + 6k̂)
b̅₁ = 3î + 2ĵ + 6k̂
Also for r̅ = 7î − 6k̂ + μ(î + 2ĵ + 2k̂). b̅₂ = î + 2ĵ + 2k̂
Let θ be the angle between the lines

Question. Write the equation of the straight line through the point (α, β, γ) and parallel to z-axis.
Answer. Any line parallel to z-axis has direction ratios proportional to 0, 0, 1.
∴ The eqn. of a line through (α, β, γ) and parallel to z-axis is

Question. Find the cartesian equation of the line which passes through the point (–2, 4, –5) and is parallel to the line

Answer. Equation of the line can be written as

Direction ratios of this line are 3, –5, 6. The required line passes through (–2, 4, –5) and its direction ratios are proportional to 3, –5, 6. So, its equation is

Question. Write the vector equation of a line passing through the point (1, –1, 2) and parallel to the line whose equation is

Answer. Vector eq. of the line passing through (1, –1, 2)

Question. If the equation of a line AB is

Find the direction cosines of a line parallel to AB.
Answer. The line AB is given by

Question. The equation of a line is given by

Write the direction cosines of a line parallel to the above line.
Answer. The given line is

Question. Write the equation of the line parallel to the line

and passing through the point (1, 2, 3).
Answer. Equation of line passing through the point (1, 2, 3) and parallel to the given line

Question. The equation of a line is

Find the direction cosines of a line parallel to this line.
Answer. The equation of given line is

Question. Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is 2î − 3ĵ + 6k̂.
Answer.

Question. Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively. 
Answer. Intercepts are a = 3, b = –4, c = 2
∴ The intercept form of the plane is

Question. Write the equation of a plane which is at a distance of 5 √3 units from origin and the normal to which is equally inclined to coordinate axes. 
Answer. Let α, β and γ be the angles made by n̅ with x,
y and z-axis, respectively.
Given α = β = γ ⇒ cos α = cos β = cos γ
⇒ l = m = n, where l, m, n are direction cosines of n̅

Question. Find the length of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 21 = 0.
Answer. Length of the 􀁁 from O(0, 0, 0) to the plane
2x –3y + 6z + 21 = 0 is

Question. Find the distance of the plane 3x – 4y + 12z = 3 from the origin.
Answer. Perpendicular distance from the origin (0, 0, 0) to the plane 3x – 4y + 12z – 3 = 0 is

Question. Write the distance of the plane 2x – y + 2z + 1 = 0 from origin.
Answer. Perpendicular distance from the origin (0, 0, 0) to the plane 2x – y + 2z + 1 = 0 is

Question. Write the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane r̅ .(î + 2ĵ − 5k̂) + 9 = 0.
Answer. e given plane is
⇒ r̅ (î + 2ĵ − 5k̂) + 9 = 0
⇒ n̅ = î + 2ĵ − 5k̂
∵ D.r’s of ⊥ to this plane are 1, 2, –5.
So, the line has direction ratios proportional to 1, 2, –5.
∴ Eq. of line through (1, 2, 3) and ⊥ to the plane is
r̅ = (î + 2ĵ + 3k̂) + λ(î + 2ĵ − 5k̂).

Question. Find the value of λ such that the line

Answer.

Short Answer Type Questions

Question. 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).
Answer. The equation of line AB is given by

⇒ x = 7μ + 3, y = 5μ + 9, z = 4
The coordinates of a general point on CD are (7μ + 3, 5μ + 9, 4)
If the line AB and CD intersect then they have a common point. So, for some values of λ and μ, we must have
4λ = 7μ + 3, 6λ – 1 = 5μ + 9, 2λ – 1 = 4
⇒ 4λ – 7μ = 3 …(i), 6λ – 5μ = 10 …(ii) and λ = 5/2                 …(iii)
Substituting λ = 5/2
in (ii), we get μ = 1
Since λ = 5/2 and μ = 1 satisfy (i), so the given lines
AB and CD intersect.

Question. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.
Answer. The equation of the line through A(3, 4, 1)

Any point on (i) is given by (2k + 3, –3k + 4, 5k + 1)
We know that the coordinates of any point on the
XZ plane are (x₁, 0, z₁).
∵ (2k + 3, –3k + 4, 5k + 1) lies on the XZ plane,
∴ –3k + 4 = 0 ⇒ k = 4/3
Thus the coordinates of the point where the line joining A and B crosses the XZ plane are

Question. Find the equation of the plane passing through the points (–1, 2, 0), (2, 2, –1) and parallel to the line

Answer. Eq. of any plane through (–1, 2, 0) is
a(x + 1) + b(y – 2) + cz = 0                               …(i)
Also it passes through (2, 2, –1)
∴ 3a + 0 · b – c = 0                                           …(ii)
Further plane (i) is parallel to the line

⇒ a = λ, b = 2λ and c = 3λ
Put these values in (i), we get
x + 2y + 3z = 3
This is the eq. of the required plane

Question. Find the equation of a plane which passes through the point (3, 2, 0) and contains the line
Answer. Equation of plane passing through (3, 2, 0) is
a(x – 3) + b(y – 2) + c(z – 0) = 0                              …(i)

Since plane contains the line, so
a(3 –3) + b(6 – 2) + c(4 – 0) = 0
⇒ 0·a + 4b + 4c = 0                             …(ii)
and a(1) + b(5) + c(4) = 0
⇒ a + 5b + 4c = 0                             …(iii)
Solving (ii) & (iii), we get

⇒ a = –4λ, b = 4λ, c = –4λ
Putting values of a, b, c in (i), we get
–4λ(x – 3) + 4λ(y – 2) – 4λ(z – 0) = 0
⇒ –4x + 12 + 4y – 8 – 4z = 0
⇒ x – y + z – 1 = 0 is the required equation of plane.

Question. Find the image of the point (2, –1, 5) in the line

Also find the equation of the line joining the given point and its image. Find the length of that line-segment also.
Answer. The given line is

Any point on it is
M(10λ + 11, –4λ – 2, –11λ – 8).
Let the given point be P(2, –1, 5)
D.r’s of PM are 10λ + 11 – 2, –4λ – 2 + 1, –11λ –8 –5.
i.e., 10λ + 9, –4λ – 1, – 11λ – 13.
As line (i) is ⊥ to PM, so
(10λ + 9) · 10 + (–4λ –1) · (– 4)
+ (–11λ – 13) · (–11) = 0
⇒ (100 + 16 + 121)λ + (90 + 4 + 143) = 0
⇒ 237λ + 237 = 0 ⇒ λ = –1.
∴ D.r’s of PM are –1, 3, –2 i.e., 1, –3, 2.
∴ Eq. of the line PM is

Question. Using vectors, show that the points A(–2, 3, 5) B(7, 0, –1) C(–3, –2, –5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3)
Answer.

Question. Find the perpendicular distance of the the point (1, 0, 0) from the line

Also find the coordinates of the foot of the perpendicular and the equation of the perpendicular
Answer. Any point on the given line

is R(2k + 1, –3k – 1, 8k –10)
If this is the foot of the ⊥ from P(1, 0, 0) on (i), then
(2k + 1 –1) · 2 + (–3k – 1 – 0) · (–3)
+ (8k –10 – 0) · 8 = 0
⇒ 4k + 9k + 3 + 64k – 80 = 0
⇒ 77k = 77 ⇒ k = 1.
∴ R is (3, –4, –2).
This is the required foot of perpendicular.
Also, perpendicular distance = PR

Question. Find the equation of the perpendicular from the point (3, –1, 11) to the line

Also find the coordinates of the foot of the perpendicular and the length of the perpendicular.
Answer. Any point on the given line

is R(2k, 3k + 2, 4k + 3).
If R is the foot of the ⊥ from P(3, –1, 11) on (i), then
line PR has d.r’s 2k – 3, 3k + 2 + 1, 4k + 3 –11
∴ 2(2k – 3) + 3(3k + 3) + 4(4k – 8) = 0
⇒ 29k – 29 = 0 ⇒ k = 1.
∴ D.r’s of perpendicular line are –1, 6, –4.
As it is drawn through P(3, –1, 11), so its eq. is

Question. Find the points on the line

at a distance of 5 units from the point P(1, 3, 3).
Answer.

is of the form Q(3λ – 2, 2λ – 1, 2λ + 3)                         …(i)
Now, distance PQ, where P is (1, 3, 3), is 5.
So, (3λ – 2 – 1)2 + (2λ – 1– 3)2 + (2λ + 3 – 3)2 = 52
⇒ 9λ2 + 9 – 18λ + 4λ2 + 16 – 16λ + 4λ2 = 25
⇒ 17λ2 – 34λ = 0 ⇒ 17λ(λ – 2) = 0
⇒ λ = 0 or λ = 2
Putting values of λ in (i), we get the required points
are (– 2, – 1, 3) and (4, 3, 7).

Question. Find the foot of the perpendicular from the point (0, 2, 3) on the line

Also, find the length of the perpendicular
Answer. Let Q be the foot of perpendicular from P(0, 2, 3) on

Now, coordinates of Q be (5λ – 3, 2λ + 1, 3λ – 4).
Direction ratios of PQ are 5λ – 3, 2λ + 1 – 2, 3λ – 4 – 3
or 5λ – 3, 2λ – 1, 3λ – 7
Since PQ is perpendicular to the given line,
∴ 5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0
⇒ 25λ – 15 + 4λ – 2 + 9λ – 21 = 0
⇒ 38λ = 38 􀂟 λ = 1
∴ Coordinates of Q are (2, 3, – 1)

Question. Find the equation of the line which is parallel to 2î − ĵ + 3k̂ and which passes through the point (5, –2, 4).
Answer.

Question. Find the vector and cartesian equations of the line through the point (1, 2, –4) and perpendicular to the two lines

Answer. The given lines are

Question. Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines

Answer. Let the eq. of line passing through (2, 1, 3) and perpendicular to the lines

Question. Find the value of p, so that the lines

perpendicular to each other. Also nd the equation of a line passing through a point (3, 2, –4) and parallel to line l₁.
Answer. The given lines are

Question. A line passes through (2, –1, 3) and is perpendicular to the lines

Obtain its equation in vector and cartesian form.
Answer. The given lines are

Question. Find the direction cosines of the line

Also, find the vector equation of the line through the point A (–1, 2, 3) and parallel to the given line.
Answer.

Question. The cartesian equations of a line are 6x – 2 = 3y + 1 = 2z –2. Find the direction cosines of the line. Write down the cartesian and vector equations of a line passing through (2, –1, –1) which is parallel to the given line
Answer. The cartesian eqs. of the given line are
6x – 2 = 3y + 1 = 2z –2

Question. Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the lines
Answer. Let l, m, n be the direction ratios of the line which is perpendicular to the lines

Question. Find the angle between the following pair of

Answer. Writing given lines in standard form as

Question. Find the value of λ, so that the following lines are perpendicular to each other

Answer. The given lines

Question. Find the equation of the perpendicular drawn from the point (2, 4, –1) to the line
Answer. Let the foot of perpendicular drawn from

Now, coordinates of Q be (λ – 5, 4λ – 3, – 9λ + 6)
Direction ratios of PQ be
λ – 5 – 2, 4λ – 3 – 4, – 9λ + 6 + 1
or λ – 7, 4λ – 7, –9λ + 7
Since line PQ and given line are perpendicular.
∴ λ – 7 + 4(4λ – 7) – 9(– 9λ + 7) = 0
⇒ λ – 7 + 16λ – 28 + 81λ – 63 = 0
⇒ 98λ – 98 = 0 ⇒ 􀀃λ = 1
So, direction ratios of PQ are <– 6, – 3, – 2>.

Question. Find the equation of a line, which passes through the point (1, 2, 3) and is parallel to the line
Answer.

Question. Find the shortest distance between the lines

Answer. The given lines are

Question. Find the shortest distance between the following

Answer. The given lines are

Question. Find the shortest distance between the lines whose vector equations are

Answer.

Question. Find the shortest distance between the following lines:

Answer.

Question. Find the distance between the lines l₁ and l₂ given by

Answer. Given lines are

Long Answer Type Questions

Question. Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and whose x-intercept is twice its z-intercept.
Answer. Equation of the plane passing through the line of intersection of the given plane is (x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
⇒ (2λ + 1) x + (λ + 2) y + (3 – λ) z = 4 – 5λ
This equation of the plane can be written in the intercept form as

Question. Find the vector and cartesian equations of the plane passing through the line of intersection of the planes r̅  .(2î + 2ĵ − 3k̂) = 7, r̅ · (2î + 5ĵ − 3k̂) = 9 such that the intercepts made by the plane on x-axis and z-axis are equal.
Answer. The given planes are
r̅ · (2î + 2ĵ − 3k̂) = 7 and r̅ · (2î + 5ĵ − 3k̂) = 9
⇒ 2x + 2y – 3z = 7 and 2x + 5y – 3z = 9
∴ Equation of the plane passing through their line of intersection is
(2x + 2y – 3z – 7) + λ(2x + 5y – 3z – 9) = 0
⇒ (2 + 2λ) x + (2 + 5λ) y + (– 3λ – 3)z – (7 + 9λ) = 0 …(i)
Given, x-intercept = z-intercept

⇒ λ = – 1
Put λ = –1 in (i), we get, 3y = 2
This is the cartesian eq. of the plane.
Corresponding vector eq. is r̅ · (3ĵ) = 2

Question. Find the vector equation of the plane passing through three points with position vectors î + ĵ − 2k̂, 2î − ĵ + k̂ and î + 2ĵ + k̂. Also find the coordinates of the point of intersection of this plane and the line r̅  = 3î − ĵ − k̂ + λ (2î − 2ĵ + k̂).
Answer. The given points are
P (î + ĵ − 2k̂), (1, 1, − 2), Q(2î − ĵ + k̂) , (2, −1, 1)
and R(î + 2ĵ + k̂), (1, 2, 1)
Eq. of any plane through P is
a(x – 1) + b(y – 1) + c(z + 2) = 0                             …(i)
As it is also passing through Q and R, so
a – 2b + 3c = 0                                                       … (ii)
0 · a + b + 3c = 0                                                   … (iii)
Solving (ii) and (iii), we get

∴ From (i), equation of plane is
9 (x – 1) + 3(y – 1) – (z + 2) = 0
⇒ 9x + 3y – z – 14 = 0 ⇒ r̅ . (9î + 3ĵ − k̂) = 14
Now any point on the line
r̅ = 3î − ĵ − k̂ + λ(2î − 2ĵ + k̂)

Question.

find the value of k and hence find the equation of the plane containing these lines.
Answer.

Question. Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, –3), B(–2, –3, 5) and C(5, 3, –3). 
Answer. The equation of the plane passing through
(2, 5, – 3) is
a(x – 2) + b(y – 5) + c(z + 3) = 0 …(i)
If plane (i) passes through the points (–2, –3, 5) and
(5, 3, – 3), then
– 4a – 8b + 8c = 0 ⇒ a + 2b – 2c = 0 …(ii)
and 3a – 2b + 0·c = 0 …(iii)
Solving (ii) and (iii), we get

Substituting the values of a, b, c in (i), we get
2λ(x – 2) + 3λ(y – 5) + 4λ(z + 3) = 0
⇒ 2x + 3y + 4z = 7 which is the equation of the plane passing through the given three points
Now the distance of (7, 2, 4) from this plane is given by

Question. Find the distance of the point (2, 12, 5) from the point of intersection of the line r̅ = 2î − 4ĵ + 2k̂ + λ(3î + 4ĵ + 2k̂) and the plane r̅ · (î − 2ĵ + k̂) = 0.
Answer. The given line is

Question. Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence find the distance of point P(–2, 5, 5) from the plane obtained above.
Answer. Given planes are 2x + 3y – 2z = 5 …(i)
and x + 2y – 3z = 8 …(ii)
Normal vectors of (i) and (ii) are respectively

Question. Find the vector equation of the plane that contains the lines r̅ = (î + ĵ ) + λ (î + 2ĵ − k̂) and r̅ = (î + ĵ ) + μ (−î + ĵ − 2k̂). Also find the length of perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.
Answer. The given lines are

Question. Find the distance of the point (2, 3, 4) from the line

measured parallel to the plane 3x + 2y + 2z – 5 = 0.
Answer. Given point be P(2, 3, 4) and equation of given line

Any point on the line be Q(3λ – 3, 6λ + 2, 2λ).
Direction ratios of PQ are
3λ – 3 – 2, 6λ + 2 – 3, 2λ – 4 i.e., 3λ – 5, 6λ – 1, 2λ – 4
Since PQ is parallel to the plane 3x + 2y + 2z – 5 = 0
∴ 3(3λ – 5) + 2(6λ – 1) + 2(2λ – 4) = 0
⇒ 9λ – 15 + 12λ – 2 + 4λ – 8 = 0
⇒ 25λ – 25 = 0 ⇒ λ = 1
So, coordinates of Q are (0, 8, 2)

Question. Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0 measured parallel to the line

Answer. Equation of a line through P(2, 3, 4) and parallel to given line is

⇒ x = 3λ + 2, y = 6λ + 3, z = 2λ + 4
Coordinates of any point on this line are
(3λ + 2, 6λ + 3, 2λ + 4)
and this point must lie on the plane 3x + 2y + 2z + 5 = 0.
∴ 3(3λ + 2) + 2(6λ + 3) + 2(2λ + 4) + 5 = 0
⇒ 9λ + 6 + 12λ + 6 + 4λ + 8 + 5 = 0
⇒ 25λ + 25 = 0 ⇒ λ = –1
∴ Coordinates of points Q lying on plane are
(–1, –3, 2).
Distance between the points P(2, 3, 4) and Q (–1, –3, 2) is given by

Question. From the point P(1, 2, 4) a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.
Answer. Let foot of perpendicular from P(1, 2, 4) be Q.
Then, PQ is normal to the plane. So, its direction ratios are proportional to 2, 1, – 2. Since PQ passes through P(1, 2, 4), so its equation is

Let coordinates of Q be (2λ + 1, λ + 2, 4 – 2λ).
Since it lies on the plane, therefore
2(2λ + 1) + λ + 2 – 2(4 – 2λ) + 3 = 0
⇒ 4λ + 2 + λ + 2 – 8 + 4λ + 3 = 0
⇒ 9λ – 1 = 0 ⇒ λ =1/9
So, coordinates of Q are

Question. Find the length and the foot of the perpendicular from the point P(7, 14, 5) to the plane 2x + 4y – z = 2. Also find the image of point P in the plane. 
Answer. The given plane is 2x + 4y – z = 2 …(i)
The d.r.s. of the normal to (i) are 2, 4, –1.
∴ Eq. of line ⊥ to (i) through P(7, 14, 5) is

Question. Find the equation of the plane passing through the point (3, –3, 1) and perpendicular to the line joining the points (3, 4, –1) and (2, –1, 5). Also find the coordinates of foot of perpendicular, the equation of perpendicular line and the length of perpendicular drawn from origin to the plane.
Answer. The line joining the given points
P(3, 4, –1) and Q(2, –1, 5) has direction ratios
<3 –2, 4 + 1, –1 – 5> i.e., <1, 5, –6>
The plane through (3, –3, 1) and perpendicular to the line PQ is
1(x – 3) + 5(y + 3) – 6(z – 1) = 0
⇒ x + 5y – 6z + 18 = 0 …(i)
Eq. of the line ⊥ to (i) through the origin is

Question. Show that the lines r̅ = (î + ĵ − k̂) + λ (3î − ĵ) and r̅ = (4î − k̂) + μ(2î + 3k̂) are coplanar. Also find the equation of the plane containing them.
Answer. The given lines are