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Title: BEST 3DIMENSIONAL GEOMETRY PROBLEMS
Description: The notes have been typed by myself.It includes 3D Geometry Problems which includes basics and High level problems.
Description: The notes have been typed by myself.It includes 3D Geometry Problems which includes basics and High level problems.
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Exercise 11
...
Answer
Let direction cosines of the line be l, m, and n
...
Answer
Let the direction cosines of the line make an angle α with each of the coordinate axes
...
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes,
are
Question 3:
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
Answer
If a line has direction ratios of −18, 12, and −4, then its direction cosines are
...
Answer
The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7)
...
The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i
...
, −3, −5, and −3
...
e
...
It can be seen that the direction ratios of BC are −2 times that of AB i
...
, they are
proportional
...
Since point B is common to both AB and BC, points A, B,
and C are collinear
...
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Question 5:
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (−
1, 1, 2) and (− 5, − 5, − 2)
Answer
The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2)
...
e
...
Therefore, the direction cosines of AB are
The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i
...
, −4, −6, and −4
...
e
...
Therefore, the direction cosines of AC are
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vidhyarjan
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(One Km from ‘Welcome’ Metro Station)
Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Exercise 11
...
Answer
Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each
other, if l1l2 + m1m2 + n1n2 = 0
(i) For the lines with direction cosines,
and
, we obtain
and
, we obtain
Therefore, the lines are perpendicular
...
(iii) For the lines with direction cosines,
and
, we obtain
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Thus, all the lines are mutually perpendicular
...
Answer
Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line
joining the points, (0, 3, 2) and (3, 5, 6)
...
e
...
The direction ratios, a2, b2, c2, of CD are (3 − 0), (5 − 3), and (6 −2) i
...
, 3, 2, and 4
...
Question 3:
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the
points (−1, −2, 1), (1, 2, 5)
...
The directions ratios, a1, b1, c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i
...
, −2, −4, and
−4
...
e
...
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Question 4:
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to
the vector
...
Therefore, the position
vector through A is
It is known that the line which passes through point A and parallel to
is given by
is a constant
...
Question 5:
Find the equation of the line in vector and in Cartesian form that passes through the
point with position vector
and is in the direction
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Class XII
Chapter 11 – Three Dimensional Geometry
It is known that a line through a point with position vector
Maths
and parallel to
is given by
the equation,
This is the required equation of the line in vector form
...
Question 6:
Find the Cartesian equation of the line which passes through the point
(−2, 4, −5) and parallel to the line given by
Answer
It is given that the line passes through the point (−2, 4, −5) and is parallel to
The direction ratios of the line,
, are 3, 5, and 6
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Therefore the equation of the required line is
Question 7:
The Cartesian equation of a line is
...
Answer
The Cartesian equation of the line is
The given line passes through the point (5, −4, 6)
...
This means that the line is in the direction of vector,
It is known that the line through position vector
and in the direction of the vector
is
given by the equation,
This is the required equation of the given line in vector form
...
Answer
The required line passes through the origin
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
The line is parallel to the vector given by the equation,
The equation of the line in vector form through a point with position vector
to
and parallel
is,
The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given
by,
Therefore, the equation of the required line in the Cartesian form is
Question 9:
Find the vector and the Cartesian equations of the line that passes through the points (3,
−2, −5), (3, −2, 6)
...
Since PQ passes through P (3, −2, −5), its position vector is given by,
The direction ratios of PQ are given by,
(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is
The equation of PQ in vector form is given by,
The equation of PQ in Cartesian form is
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e
...
The angle between the given pairs of lines is given by,
The given lines are parallel to the vectors,
and
,
respectively
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Class XII
Chapter 11 – Three Dimensional Geometry
(ii) The given lines are parallel to the vectors,
Maths
and
,
respectively
...
Answer
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Let
Chapter 11 – Three Dimensional Geometry
and
Maths
be the vectors parallel to the pair of lines,
, respectively
...
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Answer
The given equations can be written in the standard form as
and
The direction ratios of the lines are −3,
, 2 and
respectively
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Class XII
Chapter 11 – Three Dimensional Geometry
Thus, the value of p is
Maths
...
Answer
The equations of the given lines are
and
The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively
...
Question 14:
Find the shortest distance between the lines
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Class XII
Chapter 11 – Three Dimensional Geometry
Therefore, the shortest distance between the two lines is
Maths
units
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Substituting all the values in equation (1), we obtain
Since distance is always non-negative, the distance between the given lines is
units
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Class XII
Chapter 11 – Three Dimensional Geometry
Comparing the given equations with
and
Maths
, we obtain
Substituting all the values in equation (1), we obtain
Therefore, the shortest distance between the two given lines is
units
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Class XII
Chapter 11 – Three Dimensional Geometry
It is known that the shortest distance between the lines,
Maths
and
, is
given by,
For the given equations,
Substituting all the values in equation (3), we obtain
Therefore, the shortest distance between the lines is
units
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Exercise 11
...
(a)z = 2 (b)
(d)5y + 8 = 0
(c)
Answer
(a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)
The direction ratios of normal are 0, 0, and 1
...
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the
origin is 2 units
...
∴
Dividing both sides of equation (1) by
, we obtain
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Therefore, the direction cosines of the normal are
normal from the origin is
and the distance of
units
...
Dividing both sides of equation (1) by
, we obtain
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin
...
(d) 5y + 8 = 0
⇒ 0x − 5y + 0z = 8 … (1)
The direction ratios of normal are 0, −5, and 0
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin
...
Question 2:
Find the vector equation of a plane which is at a distance of 7 units from the origin and
normal to the vector
...
Question 3:
Find the Cartesian equation of the following planes:
(a)
(b)
(c)
Answer
(a) It is given that equation of the plane is
For any arbitrary point P (x, y, z) on the plane, position vector
is given by,
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(b)
For any arbitrary point P (x, y, z) on the plane, position vector
Substituting the value of
is given by,
in equation (1), we obtain
This is the Cartesian equation of the plane
...
Question 4:
In the following cases, find the coordinates of the foot of the perpendicular drawn from
the origin
...
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Dividing both sides of equation (1) by
, we obtain
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin
...
Therefore, the coordinates of the foot of the perpendicular are
(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,
y1, z1)
...
Dividing both sides of equation (1) by 5, we obtain
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The coordinates of the foot of the perpendicular are given by
(ld, md, nd)
...
… (1)
The direction ratios of the normal are 1, 1, and 1
...
The coordinates of the foot of the perpendicular are given by
(ld, md, nd)
...
⇒ 0x − 5y + 0z = 8 … (1)
The direction ratios of the normal are 0, −5, and 0
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Dividing both sides of equation (1) by 5, we obtain
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin
...
Therefore, the coordinates of the foot of the perpendicular are
Question 5:
Find the vector and Cartesian equation of the planes
(a) that passes through the point (1, 0, −2) and the normal to the plane is
...
Answer
(a) The position vector of point (1, 0, −2) is
The normal vector
perpendicular to the plane is
The vector equation of the plane is given by,
is the position vector of any point P (x, y, z) in the plane
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
This is the Cartesian equation of the required plane
...
Therefore, equation (1) becomes
This is the Cartesian equation of the required plane
...
(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)
(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)
Answer
(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3)
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Since A, B, C are collinear points, there will be infinite number of planes passing through
the given points
...
Therefore, a plane will pass through the points A, B, and C
...
Question 7:
Find the intercepts cut off by the plane
Answer
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Therefore, for the given equation,
Thus, the intercepts cut off by the plane are
...
Answer
The equation of the plane ZOX is
y=0
Any plane parallel to it is of the form, y = a
Since the y-intercept of the plane is 3,
∴a=3
Thus, the equation of the required plane is y = 3
Question 9:
Find the equation of the plane through the intersection of the planes
and
and the point (2, 2, 1)
Answer
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Therefore, this point will satisfy equation
(1)
...
Question 10:
Find the vector equation of the plane passing through the intersection of the planes
and through the point (2, 1, 3)
Answer
The equations of the planes are
The equation of any plane through the intersection of the planes given in equations (1)
and (2) is given by,
, where
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Therefore, its position vector is given by,
Substituting in equation (3), we obtain
Substituting
in equation (3), we obtain
This is the vector equation of the required plane
...
The plane in equation (1) is perpendicular to
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Since the planes are perpendicular,
Substituting
in equation (1), we obtain
This is the required equation of the plane
...
Answer
The equations of the given planes are
It is known that if
and
are normal to the planes,
and
and
, then the
angle between them, Q, is given by,
Here,
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(a)
(b)
(c)
(d)
(e)
Answer
The direction ratios of normal to the plane,
, are a1, b1, c1 and
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Therefore, the given planes are not perpendicular
...
The angle between them is given by,
(b) The equations of the planes are
Here,
and
and
Thus, the given planes are perpendicular to each other
...
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(d) The equations of the planes are
Here,
and
and
∴
Thus, the given lines are parallel to each other
...
∴
Therefore, the given lines are not parallel to each other
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Question 14:
In the following cases, find the distance of each of the given points from the
corresponding given plane
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
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Answer
Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1)
...
The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and
(−1 + 1) = 0
OA is perpendicular to BC, if a1a2 + b1b2 + c1c2 = 0
∴ a1a2 + b1b2 + c1c2 = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0
Thus, OA is perpendicular to BC
...
Answer
It is given that l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually
perpendicular lines
...
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∴l2 + m2 + n2 = 1 … (5)
It is known that,
∴
Substituting the values from equations (5) and (6) in equation (4), we obtain
Thus, the direction cosines of the required line are
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Answer
The angle Q between the lines with direction cosines, a, b, c and b − c, c − a,
a − b, is given by,
Thus, the angle between the lines is 90°
...
Answer
The line parallel to x-axis and passing through the origin is x-axis itself
...
Therefore, the coordinates of A are given by (a, 0, 0), where
a ∈ R
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Question 5:
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
...
The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4
The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8
It can be seen that,
Therefore, AB is parallel to CD
...
sQuestion 6:
If the lines
and
are perpendicular, find the value
of k
...
It is known that two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are
perpendicular, if a1a2 + b1b2 + c1c2 = 0
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Question 7:
Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the
plane
Answer
The position vector of the point (1, 2, 3) is
, are 1, 2, and −5
The direction ratios of the normal to the plane,
and the normal vector is
The equation of a line passing through a point and perpendicular to the given plane is
given by,
Question 8:
Find the equation of the plane passing through (a, b, c) and parallel to the plane
Answer
Any plane parallel to the plane,
, is of the form
The plane passes through the point (a, b, c)
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Class XII
Chapter 11 – Three Dimensional Geometry
Substituting
Maths
in equation (1), we obtain
This is the vector equation of the required plane
...
Answer
The given lines are
It is known that the shortest distance between two lines,
and
, is
given by
Comparing
to equations (1) and (2), we obtain
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Question 10:
Find the coordinates of the point where the line through (5, 1, 6) and
(3, 4, 1) crosses the YZ-plane
Answer
It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,
y2, z2), is
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k)
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Class XII
Chapter 11 – Three Dimensional Geometry
Therefore, the required point is
Maths
...
Answer
It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,
y2, z2), is
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k)
...
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Answer
It is known that the equation of the line through the points, (x1, y1, z1) and (x2, y2, z2), is
Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is
given by,
Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5)
...
e
...
Question 13:
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to
each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0
...
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Question 14:
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane
, then find the value of p
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
The equation of the given plane is
It is known that the perpendicular distance between a point whose position vector is
and the plane,
is given by,
and d
Here,
Therefore, the distance between the point (1, 1, p) and the given plane is
Similarly, the distance between the point (−3, 0, 1) and the given plane is
It is given that the distance between the required plane and the points, (1, 1, p) and
(−3, 0, 1), is equal
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Question 15:
Find the equation of the plane passing through the line of intersection of the planes
and
and parallel to x-axis
...
The required plane is parallel to x-axis
...
The direction ratios of x-axis are 1, 0, and 0
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
Therefore, its Cartesian equation is y − 3z + 6 = 0
This is the equation of the required plane
...
Answer
The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively
...
Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3)
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Answer
The equations of the given planes are
The equation of the plane passing through the line intersection of the plane given in
equation (1) and equation (2) is
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The Cartesian equation of this plane can be obtained by substituting
in
equation (3)
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Answer
The equation of the given line is
The equation of the given plane is
Substituting the value of
from equation (1) in equation (2), we obtain
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The point is (−1, −5, −10)
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Answer
Let the required line be parallel to vector
given by,
The position vector of the point (1, 2, 3) is
The equation of line passing through (1, 2, 3) and parallel to
is given by,
The equations of the given planes are
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Therefore, the normal to
the plane of equation (2) and the given line are perpendicular
...
in equation (1), we obtain
This is the equation of the required line
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
The position vector of the point (1, 2, − 4) is
The equation of the line passing through (1, 2, −4) and parallel to vector
is
The equations of the lines are
Line (1) and line (2) are perpendicular to each other
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From equations (4) and (5), we obtain
∴Direction ratios of
Substituting
are 2, 3, and 6
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vidhyarjan
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(One Km from ‘Welcome’ Metro Station)
Class XII
Chapter 11 – Three Dimensional Geometry
Maths
It can be seen that the given planes are parallel
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Hence, the correct answer is D
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5y + 10z = 6 are
(A) Perpendicular (B) Parallel (C) intersect y-axis
(C) passes through
Answer
The equations of the planes are
2x − y + 4z = 5 … (1)
5x − 2
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Class XII
Chapter 11 – Three Dimensional Geometry
Maths
∴
Therefore, the given planes are parallel
...
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Title: BEST 3DIMENSIONAL GEOMETRY PROBLEMS
Description: The notes have been typed by myself.It includes 3D Geometry Problems which includes basics and High level problems.
Description: The notes have been typed by myself.It includes 3D Geometry Problems which includes basics and High level problems.