Three Dimensional Geometry Class 12 Mathematics Exam Questions

Please refer to Three Dimensional Geometry Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 12 Mathematics Exam Questions Three Dimensional Geometry

Class 12 Mathematics students should read and understand the important questions and answers provided below for Three Dimensional Geometry which will help them to understand all important and difficult topics.

Very Short Answer Type Questions

Question. Find the length of the intercept, cut off by the plane 2x + y – z = 5 on the x-axis
Answer.

Three Dimensional Geometry Class 12 Mathematics Exam Questions

Question. Find the direction cosines of the line

Answer.

Question. The equation of a line are 5x – 3 = 15y + 7 = 3 –10z. Write the direction cosines of the line.
Answer. The given line is 5x – 3 = 15y + 7 = 3 –10z

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 y be the angles made by n̅ with x, y and z-axis, respectively.
Given α = β = y ⇒ cos a = cos β = cos y
⇒ l = m = n, where l, m, n are direction cosines of n̅ .

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) and parallel to the line

r̅ = (î – ĵ + 2k̂ + λ (î + 2ĵ – 2k̂)

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. 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. let, n̅ 2î -3 ĵ + 6k̂

Question. Find the vector equation of a line which passes through the points (3, 4, –7) and (1, –1, 6).
Answer. Vector equation of a line passes through the points (3, 4, –7) and (1, –1, 6) is given by

Question. Find the distance between the planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20.
Answer. Equation of both the planes can be written as 2x – y + 2z = 5 and 2x – y + 2z = 8.
Distance between both the planes

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. The 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 – 2k̂)

Short Answer Type Questions -I

Question. Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
Answer. Vector equation of the line passing through (1, 2, –1) and parallel to the line
5x – 25 = 14 – 7y = 35z

Question. Find the value of k so that the lines x = – y = kz and x – 2 = 2y + 1 = –z + 1 are perpendicular to each other.
Answer. The given lines are

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 distance between the point (–1, –5, –10) and the point of intersection of the

Answer. Equation of given line is

⇒ x = 3k + 2, y = 4k –1, z = 12k + 2
Since point (3k + 2, 4k –1, 12k + 2) lie on plane
x – y + z = 5
∴ 3k + 2 – 4k + 1 + 12 k + 2 = 5
⇒ 11k = 0 ⇒ k = 0
∴ Point is (2, –1, 2)

Question. Find the coordinates of the point where the line through (–1, 1, –8) and (5, –2, 10) crosses the ZX-plane.
Answer. The equation of line through A(–1, 1, –8) and

Any point on (i) is given by (6k – 1, –3k + 1, 18k – 8).
We know that the coordinates of any point on the ZX-plane is given by (x, 0, z).
∴ –3k + 1 = 0 ⇒ k = 1/3
Thus the coordinates of the point where the line joining A and B crosses the ZX-plane are

Question. Find the perpendicular distance of 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 value of λ, so that the lines

at right angles. Also, find whether the lines are intersecting or not.
Answer. The given lines are

Question.

Answer. Any point on the line

is (k + 2, 3k + 4, 5k + 6)
For lines (i) and (ii) to intersect, we must have
3r – 1 = k + 2, 5r – 3 = 3k + 4, 7r – 5 = 5k + 6

Question. The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, –2) is 4. Find its z-coordinate.
Answer. Given that P(2, 2, 1) and Q(5, 1, –2)

Let the point R on the line PQ, divides the line in the ratio k : 1 and x-coordinate of point R on the line is 4.
So, by section formula

Question. A plane makes intercepts –6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.
Answer. Given that –6, 3, 4 are intercepts on x, y and z-axes respectively.

⇒ –2x + 4y + 3z –12 = 0
∴ Length of the perpendicular from origin to the
plane – 2x + 4y + 3z – 12 = 0 is

Short Answer Type Questions -II

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. a̅ = – 2î + 3ĵ + 5k̂, b̅ = 7î – k̂
∴ Equation of line joining A and B is,

Question. Find the vector and cartesian equations of the line through the point (1, 2, –4) and perpendicular to the two lines
r̅ = (8î – 19ĵ + 10k̂) + λ (3î – 16ĵ + 7k̂); and
r̅ = (15î + 29ĵ + 5k̂) + μ(3î – 8ĵ + 5k̂)
Answer. The given lines are

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)

Question. Show that the lines
r̅ = 3î + 2ĵ – 4k̂ + λ (î + 2ĵ – 2k̂);
r̅ = 3î + 2ĵ + μ (3î + 2ĵ + 6k̂) are intersecting.
Hence find their point of intersection.
Answer. The given lines are

∴ The given lines intersect and their point of intersection is (–1, –6, –12) .

Question. Find the shortest distance between the following lines :
r̅ = 2î – 5ĵ + k̂ + λ (3î + 2ĵ + 6k̂); and
r̅ = 7î – 6k̂ + μ (î + 2ĵ + 2k̂)
Answer. The given lines are

Question. Find the vector equation of the plane through the points (2, 1, –1) and (–1, 3, 4) and perpendicular to the plane x – 2y + 4z = 10.
Answer. Let the eq. of the plane through (2, 1, –1) be
a(x – 2) + b(y – 1) + c(z + 1) = 0 . ..(i)
Also, point (–1, 3, 4) lies on it.
∴ –3a + 2b + 5c = 0 …(ii)
Also (i) is ^ to the plane x – 2y + 4z = 10
∴ a · 1 + b(–2) + c · 4 = 0
⇒ a – 2b + 4c = 0 …(iii)
Solving (ii) and (iii), we get

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

Question. Find the shortest distance between the following lines

Answer.

Question. Find the shortest distance between the two lines whose vector equations are
r̅ = (î + 2ĵ + 3k̂) + λ (î – 3ĵ + 2k̂); and
r̅ = (4î + 5ĵ + 6k̂) + μ(2î – 3ĵ + k̂)
Answer. Here, the lines are

The shortest distance between the lines is given by

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 unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2î – ĵ + k̂, + î + ĵ + 2k̂; and 2î + 3k̂ respectively
Answer. Here the position vectors of A, B and C are

Question. Find the image of the point having position vector î – 3ĵ + 4k̂, in the plane r̅ . (2î – ĵ + k̂) + 3 = 0
Answer. Let P be the point with position vector

Any point M on the line is (2λ + 1, –λ + 3, λ + 4)
This lies on the plane (i).
∴ 2(2λ + 1) – (–λ + 3) + (λ + 4) + 3 = 0

⇒ 6l + 6 = 0 ⇒ l = –1.
∴ Coordinates of point M are (–1, 4, 3).
Let the image of P in the plane (i) be Q(a, b, g), then M
will be the midpoint of PQ.

Question. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.
Answer. The eq. of the line through A(3, 4, 1) and

Question. Show that the lines

Answer. The given lines are

= 3(12 + 5) + 3(12 + 35) + 8(4 – 28)
= 51 + 141 – 192 = 0 = R.H.S.
Hence the given lines are coplanar.

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.

Long Answer Type Questions

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

Any point on it is (3λ + 2, 4λ – 4, 2λ + 2)
This lies on the plane r̅ = (î – 2ĵ + k̂) = 0
⇒ x – 2y + z = 0 …(ii)
∴ 3λ + 2 – 2 (4λ – 4) + 2λ + 2 = 0
⇒ –3λ + 12 = 0 ⇒ λ = 4
∴ The point of intersection of (i) and (ii) is
(3 × 4 + 2, 4 × 4 – 4, 2 × 4 + 2) = (14, 12, 10)
Its distance from the point (2, 12, 5)

Question. Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P(5, 4, 2) to the line, r̅ = – î + 3ĵ + k̂ + λ (2î + 3ĵ – k̂);
Also find the image of P in this line.
Answer. The given line is

Any point Q on (i) is (2λ – 1, 3λ + 3, – λ + 1)
Also, the given point is P(5, 4, 2).
Now d.r’s of the line PQ are
(2λ – 1 – 5, 3l + 3 – 4, –l + 1 – 2) = (2λ – 6, 3λ – 1, –l –1).
For PQ to be ^ to (i), we must have
(2λ – 6). 2 + (3λ – 1) . 3 + (– λ – 1). (–1) = 0
⇒ 14λ – 14 = 0 ⇒ λ = 1
∴ Q is (1, 6, 0)
which is the foot of ⊥ from P on line (i).

Question.

Also, find the coordinates of the point of intersection.
Find the equation of the plane containing the two lines.
Answer. Any point on the line

is (k + 2, 4k + 3, 2k + 4)
For lines (i) and (ii) to intersect, we must have
r + 2 = k + 2, 3r + 2 = 4k + 3, r + 3 = 2k + 4
On solving these, we get, r = k = –1
∴ Lines (i) and (ii) intersect and their point of intersection is (1, –1, 2)
Now, required equation of plane is

⇒ (x – 2)(6 – 4) – (y – 2)(2 – 1) + (z – 3)(4 – 3) = 0
⇒ 2x – 4 – y + 2 + z – 3 = 0 ⇒ 2x – y + z – 5 = 0

Question. Find the vector equation of a line passing through the point (2, 3, 2) and parallel to the line r̅ = ( – 2î + 3ĵ) + λ (2î – 3ĵ + 6k̂ Also, find the distance between these two lines.
Answer. Vector equation of a line passing through (2, 3, 2) and parallel to the line

Question. Find the vector and cartesian equations of a plane containing the two lines
r̅ = 2î + ĵ + 3k̂ + λ (î + 2ĵ + 5k̂); and
r̅ = 3î + 3ĵ + 2k̂ + μ (3î + 2ĵ + 5k̂)
Also show that the line
r̅ = 2î + 5ĵ + 2k̂ + p (3î – 2ĵ + 5k̂);
Answer. The given lines are