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Three-dimensional geometry bridges imagination and mathematics, giving form to the invisible and precision to the abstract. Three-dimensional geometry is an important topic of Class 12 Maths, which helps understand the geometrical relationship between points, lines, and planes in 3-D space. In this chapter, we study basic geometric concepts and learn about new methods to solve problems in real-life scenarios like finding the coordinates of an aeroplane, satellite, etc. Class 12 NCERT solutions OF Three-dimensional geometry cover all topics, including the distance formula in 3D, direction cosines and direction ratios, equations of lines and planes, and many more. We will also learn to calculate the shortest distance between a point and a plane using the necessary formulae and detailed step-by-step solutions.
In three dimensions, every angle is a perspective, and every shape tells a story in space. NCERT Solutions for Class 12 Maths Chapter 11 Three-Dimensional Geometry, prepared by experts at Careers360, offers detailed informational study material for students preparing for the CBSE Class 12 board exam according to the latest CBSE Syllabus. For syllabus, notes, and PDF, refer to this link: NCERT.
Students who wish to access the Class 12 Maths Chapter 11 NCERT Solutions can click on the link below to download the complete solution in PDF.
Distance Formula:
The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by:
The distance between a point A(x, y, z) and the origin O(0, 0, 0) is given by:
Section Formula: The coordinates of the point R, which divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) internally or externally in the ratio m:n, are given by:
Internal Division:
External Division:
Midpoint Formula: The coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are:
Coordinates of Centroid of a Triangle: Given the vertices (x1, y1), (x2, y2), and (x3, y3) of a triangle, the coordinates of the centroid are:
Incentre of a Triangle: Given the vertices (x1, y1), (x2, y2), and (x3, y3) of a triangle, the coordinates of the incenter are:
Centroid of a Tetrahedron: Given the vertices (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), and (x4, y4, z4) of a tetrahedron, the coordinates of the centroid are:
Direction Cosines of a Line: If a directed line OP makes angles α, β, and γ with the positive X-axis, Y-axis, and Z-axis, respectively, then the direction cosines l, m, and n are:
l = cos α, m = cos β, n = cos γ
Also, the sum of squares of direction cosines is always 1:
Direction Ratios of a Line: Direction ratios of a line are denoted as a, b, and c. They are proportional to the direction cosines:
Perpendicular and Parallel Lines: Two lines are perpendicular if: a1a2 + b1b2 + c1c2 = 0
Two lines are parallel if:
Projection of a Line Segment on a Line: Given points P(x1, y1, z1) and Q(x2, y2, z2) and a line with direction cosines l, m, n, the projection of PQ on the line is:
Equation of a Plane: A plane in 3-D space can be represented in various forms:
Planes Parallel to Axes: Planes parallel to the X-axis, Y-axis, and Z-axis are represented as:
Plane Parallel to X-axis: by + cz + d = 0
Plane Parallel to Y-axis: ax + cz + d = 0
Plane Parallel to Z-axis: ax + by + d = 0
Angle between Two Lines:
Shortest distance between skew lines:
Cartesian form:
The shortest distance between the lines
is
Distance between parallel lines:
If
Class 12 Maths chapter 11 solutions - Exercise: 11.1 Page number: 381 Total questions: 5 |
Question 1: If a line makes angles
Answer:
Let the direction cosines of the line be l, m, and n.
So, we have
Therefore, the direction cosines of the lines are
Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer:
If the line is making an equal angle with the coordinate axes. Then,
Let the common angle made be
Therefore, we can write;
And as we know the relation,
or
Thus the direction cosines of the line are
Question 3: If a line has the direction ratios –18, 12, – 4, then what are its direction cosines?
Answer:
Given a line has direction ratios of -18, 12, – 4, then its direction cosines are;
A line having direction ratio -18 has direction cosine:
A line having direction ratio 12 has direction cosine:
A line having direction ratio -4 has direction cosine:
Thus, the direction cosines are
Question 4: Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.
Answer:
We have the points, A (2, 3, 4), B (– 1, – 2, 1), C (5, 8, 7);
And as we can find the direction ratios of the line joining the points
The direction ratios of AB are
i.e.,
The direction ratios of BC are
i.e.,
We can see that the direction ratios of AB and BC are proportional to each other and are -2 times.
Hence, points A, B, and C are collinear.
Answer:
Given the vertices of the triangle
Finding each side direction ratios;
Therefore, its direction cosines values are;
Similarly, for side BC;
Therefore, its direction cosines values are;
Therefore, its direction cosines values are;
Class 12 Maths chapter 11 solutions - Exercise: 11.2 Page number: 389-390 Total questions: 15 |
Question 1: Show that the three lines with direction cosines
Answer:
Given direction cosines of the three lines;
And we know that two lines with direction cosines
Hence, we will check each pair of lines:
Lines
Lines
Lines
Thus, we have all lines that are mutually perpendicular to each other.
Answer:
We have given points where the line is passing through it;
Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and the line joining the points (0, 3, 2) and (3, 5, 6) is CD.
So, we will find the direction ratios of the lines AB and CD;
Direction ratios of AB are
Direction ratios of CD are
Now, lines AB and CD will be perpendicular to each other if
Therefore, AB and CD are perpendicular to each other.
Answer:
We have given points where the line is passing through them;
Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and the line joining the points (– 1, – 2, 1) and (1, 2, 5) is CD.
So, we will find the direction ratios of the lines AB and CD;
Direction ratios of AB are
Direction ratios of CD are
Now, lines AB and CD will be parallel to each other if
Therefore, we have now;
Hence, we can say that AB is parallel to CD.
Question 4: Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector
Answer:
It is given that the line is passing through A (1, 2, 3) and is parallel to the vector
We can easily find the equation of the line which passes through the point A and is parallel to the vector
So, we have now,
Thus, the required equation of the line.
Answer:
Given that the line is passing through the point with position vector
And we know the equation of the line which passes through the point with the position vector
So, this is the required equation of the line in the vector form.
Eliminating
Hence, this is the required equation of the line in Cartesian form.
Answer:
Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the
The direction ratios of the line,
So, the required line is parallel to the above line.
Therefore, we can take the direction ratios of the required line as 3k, 5k, and 6k, where k is a non-zero constant.
And we know that the equation of line passing through the point
Therefore, we have the equation of the required line:
or
The required line equation.
Question 7: The cartesian equation of a line is
Answer:
Given the Cartesian equation of the line;
Here, the given line is passing through the point
So, we can write the position vector of this point as;
And the direction ratios of the line are 3, 7, and 2.
This implies that the given line is in the direction of the vector,
Now, we can easily find the required equation of the line:
As we know that the line passing through the position vector
So, we get the equation.
This is the required equation of the line in the vector form.
Question 8: Find the angle between the following pairs of lines:
(i)
(ii)
Answer:
(i) To find the angle A between the pair of lines
We have two lines :
The given lines are parallel to the vectors
where
Then we have
and
Therefore, we have;
or
(ii) To find the angle A between the pair of lines
We have two lines :
The given lines are parallel to the vectors
where
Then we have
and
Therefore, we have;
or
Question 9: Find the angle between the following pair of lines:
(i)
(ii)
Answer:
(i) Given lines are;
So, we have two vectors
To find the angle A between the pair of lines
Then we have
and
Therefore, we have;
or
(ii) Given lines are;
So, we have two vectors
To find the angle A between the pair of lines
Then we have
and
Therefore, we have;
or
Question 10: Find the values of p so that the lines
Answer:
First, we have to write the given equation of lines in the standard form;
Then we have the direction ratios of the above lines as;
Two lines with direction ratios
Thus, the value of p is
Question 11: Show that the lines
Answer:
First, we have to write the given equation of lines in the standard form;
Then we have the direction ratios of the above lines as;
Two lines with direction ratios
Therefore, the two lines are perpendicular to each other.
Question 12: Find the shortest distance between the lines
Answer:
So given equation of lines;
Now, we can find the shortest distance between the lines
Now, comparing the values from the equation, we obtain
Then calculating
So, substituting the values now in the formula above we get;
Therefore, the shortest distance between the two lines is
Question 13: Find the shortest distance between the lines
Answer:
We have given two lines:
Calculating the shortest distance between the two lines,
by the formula
Now, comparing the given equations, we obtain
Then, calculating the determinant
Now, calculating the denominator,
So, we will substitute all the values in the formula above to obtain,
Since distance is always non-negative, the distance between the given lines is
Question 14: Find the shortest distance between the lines whose vector equations are
Answer:
Given two equations of line
So, we will apply the distance formula for knowing the distance between two lines
After comparing the given equations, we obtain
Then, calculating the determinant value numerator.
That implies,
Now, after substituting the value in the above formula, we get,
Therefore,
Question 15: Find the shortest distance between the lines whose vector equations are
Answer:
Given two equations of the line
So, we will apply the distance formula for knowing the distance between two lines
After comparing the given equations, we obtain
Then, calculating the determinant value numerator.
That implies,
Now, after substituting the value in the above formula, we get,
Therefore,
Class 12 Maths chapter 11 solutions - Miscellaneous Exercise Page number: 390-391 Total questions: 5 |
Question 1: Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
Answer:
Given direction ratios
Thus, the angle between the lines A is given by;
Thus, the angle between the lines is
Question 2: Find the equation of a line parallel to the x-axis and passing through the origin.
Answer:
Equation of a line parallel to the x-axis and passing through the origin
So, let A be a point on the x-axis.
Therefore, the coordinates of A are given by
Now, the direction ratios of OA are
So, the equation of OA is given by,
or
Thus, the equation of the line parallel to the x-axis and passing through the origin is
Question 3: If the lines
Answer:
Given both lines are perpendicular so we have the relation;
For the two lines whose direction ratios are known,
We have the direction ratios of the lines,
Therefore, applying the formula,
Question 4: Find,the shortest distance between lines
Answer:
Given lines are;
So, we can find the shortest distance between two lines
Now, we have from the comparisons of the given equations of lines.
So,
and
Hence the shortest distance between the two given lines is 9 units.
Answer:
Given
Two straight lines in 3D whose direction cosines are (3,-16,7) and (3,8,-5)
Now, the two vectors which are parallel to the two lines are
As we know, a vector perpendicular to both vectors
A vector parallel to this vector is
Now, as we know, the vector equation of the line which passes through point p and is parallel to vector d is
Here in our question, give point p = (1,2,-4), which means the position vector of this point is
So, the required line is
Also read,
Question: Find the distance between the parallel planes
Solution:
Given Planes:
Distance between two parallel planes:
Hence, the correct answer is
Strong foundation of basic concepts: Get yourself familiar with direction cosines (l, m, n) and direction ratios (a,b,c) to answer all the three-dimensional line questions. Also, comfortable switching the equation of a line between vector and Cartesian form.
Cosine rule identity: If
Avoid confusion: Do not mix up the position vector with the direction vector. Read the question carefully and answer accordingly.
Compare the lines: Check whether two lines are parallel, intersecting, or skew using direction ratios and point substitution.
Memorise formulas: Three-dimensional geometry has many important formulas which are necessary to solve the problems. We have provided important formulas at the beginning of the article. Memorise them from time to time.
Also read,
Here are the subject-wise links for the NCERT solutions of class 12:
Given below are the class-wise solutions of class 12 NCERT:
Here are some useful links for the NCERT books and the NCERT syllabus for class 12.
To find the angle between two lines in three-dimensional space, we use the dot product formula. The formula is:
cos(θ) = (a · b) / (|a| |b|)
where a and b are the direction vectors of the two lines. This formula helps determine the angle between two lines by relating the vectors' orientation in space, and the result will give the angle between the lines.
The distance between two parallel lines in 3D space can be calculated by finding the perpendicular distance between a point on one line and the other line. If we have two parallel lines with direction vector d, and the lines pass through points P1 and P2, then the formula for the distance D is:
D = |(P2 - P1) × d| / |d|
Here, P2-P1 is the vector between any two points on the lines, and the cross product gives the area of the parallelogram formed by the two vectors. Dividing by the magnitude of the direction vector d yields the shortest distance between the lines.
To find the equation of a plane passing through three given points in 3D, let the three points be A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3). First, calculate two vectors lying in the plane, say AB = (x2 - x1, y2 - y1, z2 - z1) and AC = (x3 - x1, y3 - y1, z3 - z1). Then, find the cross product of these vectors to get the normal vector n to the plane. The equation of the plane is then given by:
n · (r - r?) = 0
where r is a point on the plane, (r - r?) is a known point (such as A ), and n is the normal vector obtained from the cross-product.
The shortest distance between skew lines (lines that are not parallel and do not intersect) is the perpendicular distance between them. To find it, we first find two points, one on each line, say P1 on line 1 and P2 on line 2 . Next, we compute the vector between these two points, P1P2. The shortest distance is given by the formula:
D = |(P1 P2) · (d1 × d2)| / |d1 × d2|
where d1 and d2 are the direction vectors of the two lines, and d1 × d2 gives a vector perpendicular to both lines. This distance represents the shortest path between the two skew lines.
The Cartesian equation of a line in 3D geometry is given by the system of equations:
(x - x0) / a = (y - y0) / b = (z - z0) / c
Here, (x0, y0, z0) is a point on the line, and (a, b, c) are the direction ratios of the line. This equation relates the coordinates of any point on the line.
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