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Chapter 28 of RD Sharma's Class 12 Mathematics textbook explores the geometry of straight lines in three-dimensional space. Exercise 28.4 Set 1 focuses on advanced problems involving lines in space, including complex relationships between lines, planes, and points. This set challenges students to apply their comprehensive understanding of 3D analytical geometry to solve sophisticated problems involving line equations, intersections, and spatial relationships.
Vector equation: r = a + λb
Parametric equations: x = x₀ + at, y = y₀ + bt, z = z₀ + ct
Cartesian equation: (x - x₀)/l = (y - y₀)/m = (z - z₀)/n
Direction cosines: cos α = l/√(l²+m²+n²), cos β = m/√(l²+m²+n²), cos γ = n/√(l²+m²+n²)
Angle between two lines: cos θ = |l₁l₂ + m₁m₂ + n₁n₂| / √[(l₁²+m₁²+n₁²)(l₂²+m₂²+n₂²)]
Condition for parallel lines: l₁/l₂ = m₁/m₂ = n₁/n₂
Condition for perpendicular lines: l₁l₂ + m₁m₂ + n₁n₂ = 0
Distance between skew lines: d = |[(a₂-a₁) · (b₁×b₂)]| / |b₁×b₂|
Solution:
Let the foot of the perpendicular drawn from P (3, -1, 11) to the line is Q, so we have to find length of PQ is general point on the line
Coordinate of Q = , direction ratios of the given line = 2,-3,4. Since PQ is the perpendicular to the given line interface.
So, the coordinates of Q are:
Distance between P and Q is given as:
So, the required distance is units
Solution:
Let us consider the foot of the perpendicular drawn from P (1,0,0) to the line is Q. So let us find the length of PQ i.e.
Coordinate of Q =
The direction ratios of the given line:
So the Coordinates of Q are as follows:
Distance between P and Q is given by:
PQ =
PQ =
PQ =
Hence, the foot of the perpendicular = (3,-4,-2);
Length of the perpendicular = units.
Solution:
Let us consider, the foot of the perpendicular drawn from A(1,0,3) to the line joining
Points B(4,7,1) and C(3,5,3) be D. The equation of the line passing through
points B(4,7,1) and C(3,5,3) is
Let
So, the direction ratio of AD is
Line AD is the perpendicular to BC so,
Hence, coordinates of D are:
=
Solution:
Given: D is the perpendicular from A(1,0,4) on BC. So,
Equation of line passing through BC is:
Coordinates of D = ( )
Direction ratios of AD is
Line AD is perpendicular to BS so,
So, coordinates of D are =
=
Solution:
Let us consider that The foot of the perpendicular drawn from P(2,3,4) to the line
is .
Equation of the line is
Let
Coordinates of Q =
So, PQ is perpendicular to the given line,
Coordinates of Q =
=
=
Distance between P and Q is given by: PQ =
=
=
Hence, perpendicular distance from (2,3,4) to the given line is units.
Solution:
Let be the foot of the perpendicular drawn from P(2,4,-1) to the line
Given line is
Coordinate of Q (General point on the line) =
Direction ratios of PQ are:
As line PQ is perpendicular to the given line, so:
Therefore, coordinates of foot of perpendicular = {-4, 1, -3}
So equation of the perpendicular PQ is :
Solution:
Let the foot of the perpendicular drawn from P(5,4,-1) to the given line is Q, so given equation of line is:
Equating the coefficients of
Coordinate of Q =
Direction ratios of line PQ are:
As line PQ is perpendicular to the given line, so:
Coordinate of Q = { }
=
=
Length of perpendicular = PQ =
=
PQ =
Solution:
Let position vector of foot of perpendicular drawn from p on be Q . So, Q is on the line
So, position vector of Q =
is the position vector of Q - position vector of p =
Here, PQ vector is perpendicular to the given line. So,
Position vector of Q = {}
=
Foot of the perpendicular =
= Position vector of Q - Position vector of P
=
=
= units
Solution:
Let Q be the perpendicular drawn from P {} on the
vector
Let the position vector of Q be :
:
= Position Vector of Q - Position Vector of P =
As PQ vector is perpendicular to the given line,
Position Vector of Q = is
Coordinates of foot of the perpendicular:
Equation of PQ is:
Solution:
Let the foot of the perpendicular drawn from (0,2,7) to the line be Q.
Given equation of the line is
Coordinate of Q = {}
Direction Ratios of PQ are
Since, PQ is perpendicular to the given line, so
Foot of the perpendicular = {}\
=
Solution:
Let the foot of perpendicular from P (1,2,-3) to the line be Q.
Given the equation of line is
Coordinates of Q are {}
Direction Ratios of PQ are: =
Let PQ be the perpendicular to the given line, so
Coordinate of the perpendicular:
Solution:
Equation of line AB is
Coordinate of point D = {}
Direction ratios of CD =
=
As line CD is perpendicular to the line AB, so
Coordinate of D = {}
= {}
= (-1,2,-5)
Equation of CD is
Solution:
Let P = (2,4,-1)
In order to find the distance we need to find a point Q on the line. We see that line is passing through
the point Q(-5,-3,6). So, let's take this point as the required point.
The line is also parallel to the vector
Now, =
Therefore,
Solution:
Let L be the foot of the perpendicular drawn from A(1,8,4) on the line joining the points B(0,-1,3) and C(2,-3,-1).
Equation of the line passing through the points B and C is given by
Let position vector of L be,
Then, = Position vector of L - Position vector of A
Since, AL vector is perpendicular to the given line
which is parallel to
Therefore,
Putting value of lambda in Equation 1, we get:
So, coordinates of foot of the perpendicular are
Question 1. Find the shortest distance between the lines (x-1)/2 = (y+1)/3 = (z-2)/4 and (x-2)/3 = (y-1)/4 = z/5.
Question 2. Calculate the shortest distance between the skew lines x = 1 + 2t, y = 2 - t, z = 3t and x = 2 + 3s, y = 1 + s, z = 4 - 2s.
Question 3. Determine the shortest distance between the lines (x-1)/1 = (y-2)/2 = (z-3)/3 and x/2 = y/3 = z/4.
Question 4. Find the shortest distance between the lines x = 3 + 2t, y = 1 - t, z = 2t and x = 1 + s, y = 2 + 2s, z = 3 - s.
Question 5. Calculate the shortest distance between the lines (x-2)/3 = y/4 = (z+1)/5 and x/1 = (y-3)/2 = (z+2)/3.
Question 6. Find the shortest distance between the line passing through (1, 2, 3) and (2, 4, 5) and the line with direction ratios 2, -1, 2 passing through (0, 1, -1).
Question 7. Determine the shortest distance between the lines x = at, y = bt, z = ct and x = p + rs, y = q + ss, z = r + ts, where a, b, c, r, s, and t are constants.
Question 8. Calculate the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-4)/3 = (y-5)/4 = (z-6)/5.
Question 9. Find the shortest distance between the lines x = 1 + 2t, y = 2 + 3t, z = 3 + 4t and x = 2 - t, y = 3 - 2t, z = 4 - 3t.
Question 10. Determine the shortest distance between the line passing through the points (1, -1, 2) and (3, 2, -1) and the line passing through (0, 1, -1) and (2, 3, 1).
Also Read,
- Straight Line in Space
- Class 12 RD Sharma Mathematics Solutions- Chapter 28 The Straight Line in Space - Exercise 28.3
- Class 12 RD Sharma Mathematics Solutions- Chapter 28 The Straight Line in Space - Exercise 28.5
- Class 12 RD Sharma Mathematics Solutions- Chapter 28 The Straight Line in Space - Exercise 28.1| Set 1
Exercise 28.4 Set 1 in RD Sharma's chapter on The Straight Line in Space represents the pinnacle of complexity in dealing with three-dimensional geometry of lines. This exercise set is meticulously designed to challenge students with the most advanced and intricate problems, pushing them to apply their knowledge in complex spatial scenarios. The problems likely encompass a wide range of applications, from determining the shortest distance between skew lines to finding intersections of lines with planes and other lines in space. Students are expected to demonstrate not only technical proficiency in manipulating line equations and vectors but also a deep understanding of spatial relationships and geometric properties in 3D.
