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NCERT exemplar Class 12 Maths solutions chapter 11 Three Dimensional Geometry - If you observe the world around you, you see everything in three dimensions! Even a tiny strand of hair has dimensions of length, width, and depth. 3D geometry refers to the mathematics of perception, direction, and shape. NCERT exemplar Class 12 Maths chapter 11 solutions initiates that there is always a requirement of three parameters to work with the comprehensive concepts of three-dimensional geometry. NCERT exemplar solutions For Class 12 Maths chapter 11 would help you out in any three-dimensional Mathematical problem that you're stuck on and get you back in your study loop. NCERT exemplar Class 12 Maths solutions chapter 11 PDF download is helpful for students to learn offline when there is a slow internet connection. Also, read - NCERT Class 12 Maths Solutions
Question:1
Answer:
Given, is inclined at 60^{0} to and at 45^{0} to
= 10 units.
We want to find the position vector of point A in space, which is nothing but
We know, there are three axes in space: X, Y, and Z.
Let OA be inclined with OZ at an angle α.
We know, directions cosines are associated by the relation:
l² + m² + n² = 1 ….(i)
In this question, direction cosines are the cosines of the angles inclined by on , , and
So,
Substituting the values of l, m, and n in equation (i),
We know the values of and , i.e. 1/2 and 1/√2 respectively.
Therefore, we get
So is given as
..........(ii)
We have,
Inserting these values of l, m and n in equation (ii),
Also ,Put
Thus, position vector of A in space
Question:2
Answer:
Given, vector =
Point = (1, -2, 3)
We can write this point in vector form as
Let ,
We must find the vector equation of the line parallel to the vector and passing through the point
We know, equation of a line passing through a point and parallel to a given vector is denoted as
Where,
In other words, we need to find
This can be achieved by substituting the values of the vectors in the above equation. We get
This can be further rearranged, upon which we get:
Thus, the require vector equation of line is
which can also be written as
Question:3
Show that the given lines, and intersect.
Also, find the point of intersection of the lines.
Answer:
We have the lines,
Let us denote these lines as L_{1}and L_{2}, such that
where
We must show that the lines L_{1}and L_{2} intersect.
To show this, let us first find any point on line L_{1} and line L_{2}
For L_{1}:
We must find the values of x, y, and z. Therefore, let us take
Take
Take
Therefore, any point on L_{1} can be represented as .
Now,
For L_{2}:
We must find the values of x, y, and z. Therefore,
Take
Take
Take
Hence, any point on line L? can be represented as (5μ + 4, 2μ + 1, μ).
If lines L_{1} and L_{2} intersect, then there exist λ and μ such that
Substituting the value of μ from equation (v) into equation (iv),
Putting this value of in eq (v),
To check, we can substitute the values of and in equation (iii), giving us:
Therefore and also satisfy equation (iii).
So, the z-coordinate from equation (i),
And the z-coordinate from equation (ii),
So, the lines intersect at the point
Therefore the lines intersect at the point (-1, -1, -1).
Question:4
Find the angle between the lines and
Answer:
Given, lines:
We are instructed to find the angle between the lines.
The line is parallel to the vector
Let
Then, we can say the line is parallel to vector
Similarly, let
Then, we can say is parallel to the vector
If we take θ as the angle between the lines, then cosine θ is:
Substituting the values of and in the above equation,
We get
Here,
Also,
Substituting the values of in equation (i) and (ii), we get
Therefore, the angle between the lines is
Question:5
Answer:
Given: A (0, -1, -1), B (4, 5, 1), C (3, 9, 4), D (-4, 4, 4).
To prove: The line passing through A and B intersects the line passing through C and D.
Proof: We know, equation of a line passing through two points (x_{1} , y_{1} , z_{1}) and (x_{2} , y_{2} , z_{2}) is:
Hence, the equation of the line passing through A (0, -1, -1) and B (4, 5,1) is:
, where x_{1} = 0, y_{1} = -1, z_{1} = -1; and x_{2} = 4, y_{2} = 5, z_{2} = 1
Let
We must find the values of x, y, and z. Therefore,
This implies that any point on the line L_{1} is (4λ, 6λ – 1, 2λ – 1).
The equation of the line passing through points C (3, 9, 4) and D (-4, 4, 4) is:
, where x_{1} = 3, y_{1} = 9, z_{1} = 4; and x_{2} = -4, y_{2} = 4, z_{2} = 4
Let
We must find the values of x, y, and z. Therefore,
This implies that any point on line L_{2} is (-7μ +3, -5μ + 9, 4).
If the lines intersect, then there must exist a value of λ and for μ, for which
From equation (iii), we get
Substituting the value of λ in equation (i),
Substituting these values of λ and μ in equation (ii),
Since the values of λ and μ satisfy eq (ii), the lines intersect.
Hence, proved that the line through A and B intersects the line through C and D.
Question:6
Answer:
Given: x = py + q, z = ry + s, and x = p’y + q’, z = r’y + s’ are perpendicular.
To Prove: pp’ + rr’ + 1 = 0.
Proof:
Let us take x = py + q and z = ry + s.
From x = py + q;
py = x - q
From z = ry + s;
ry = z - s
So,
Or, … (i)
Now, if we take x = p’y + q’ and z = r’y + s’
From x = p’y + q’;
p’y = x - q’
From z = r’y + s’;
r’y = z - s’
So,
Or,
From (i),
Line L_{1} is parallel to (from the denominators of the equation (i))
From (ii),
Line L_{2} is parallel to (from the denominators of the equation (ii))
According to the question, L_{1} and L_{2} are perpendicular.
Therefore, the dot product of the vectors should equate to 0.
Or,
(since, in vector dot product,
Or,
Therefore, the lines are perpendicular if pp’ + rr’ + 1 = 0.
Question:7
Answer:
Given, there exists a plane which perpendicularly bisects the line joining A (2, 3, 4) and B (4, 5, 8) at right angles. We must find the equation of this plane.
First, let us find the midpoint of AB.
Since the midpoint of any line is halfway between the two end points,
= (3, 4, 6).
We can represent this as a position vector,
Next, we must find the normal of the plane,
We know, the equation of the plane which perpendicularly bisects the line joining two given points is
Where,
Substituting the values in the above equation,
Upon further simplification,
Therefore, the required equation of the plane is x + y + 2z = 19.
Question:8
Answer:
Given, the plane is at a distance of from the origin, and the normal is equally inclined to coordinate axes.
We need to find the equation of this plane.
We know, the vector equation of a plane located at a distance d from the origin is represented by:
lx + my + nz = d ….(i) , where l, m and n are the direction cosines of the normal of the plane.
Since the normal is equally inclined to the coordinate axes,
Also, we know,
This means,
if we substitute the values of l, m and n in equation (i),
So,
Therefore, the required equation of the plane is x + y + z = 9.
Question:9
Answer:
Given: the line drawn from point (-2, -1, -3) meets a plane at 90^{0} at the point (1, -3, 3). We must find the equation of the plane.
Any line perpendicular to the plane is the normal.
Let the points be P (-2, -1, -3) and Q (1, -3, 3), then the line PQ is a normal to the plane.
Hence, PQ = (1 + 2, -3 + 1, 3 + 3)=> PQ = (3, -2, 6)
=> Normal to the plane =
The vector equation of a plane is represented by
Putting the obtained values in this equation,
We get,
Therefore, the required equation of the plane is 3x - 2y + 6z = 27.
Question:10
Find the equation of the plane through the points (2, 1, 0), (3, -2, -2) and (3, 1, 7).
Answer:
Given points are (2, 1, 0), (3, -2, -2), and (3, 1, 7).
We know, equation of a line passing through 3 non-collinear points (x_{1} , y_{1} , z_{1} ), (x_{2} , y_{2} , z_{2} ) and (x_{3} , y_{3} , z_{3} ) is given as:
Where, (x_{1} , y_{1} , z_{1} ) = (2, 1, 0)
(x_{2} , y_{2} , z_{2} ) = (3, -2, -2)
(x_{3} , y_{3} , z_{3} ) = (3, 1, 7)
Therefore, x_{1} = 2, y_{1} = 1, z_{1} = 0; x_{2} = 3, y_{2} = -2, z_{2} = -2; x_{3} = 3, y_{3} = 1, z_{3} = 7
Substituting these values in the line equation,
Now, since
Hence, the required equation of the plane is 7x + 3y - z = 17.
Question:11
Find the equations of the 2 lines through the origin which intersect the line at angles of each.
Answer:
Given the equation of the line, we need to find the equations of two lines through the origin which intersect the given line.
According to the theorem, equation of a line with direction ratios d_{1} = (b_{1} , b_{2} , b_{3} ) that passes through the point (x_{1} , y_{1} , z_{1} ) is expressed as:
We also know, the angle between two lines with direction ratios d_{1} and d_{2} respectively is given by:
We use these theorems to find the equations of the two lines.
Let the equation of a line be:
Given that it passes through the origin, (0, 0, 0)
Therefore, equation of both lines passing through the origin will be :
Let,
Direction ratio of the line = (2, 1, 1)
If we represent the direction ratio in terms of a position vector,
Any point on the line is given by (x, y, z). From (ii),
Hence, any point on line (ii) is
Since line (i) passes through the origin, we can say
We can represent the direction ratio in terms of position vector like:
From the theorem, we know
If we substitute the values of d? and d? from (iv) and (vi) in the above equation, and putting from the question:
Solving the numerator,
Solving the denominator,
And cos π/3 = 1/2
Substituting the values, we get
Performing cross multiplication,
Squaring both sides,
Therefore, from equation (v)
Direction ratio =
Putting μ = -1:
Direction Ratio = (2(-1) + 3, (-1) + 3, -1)
⇒ Direction Ratio = (-2 + 3, -1 + 3, -1)
⇒ Direction Ratio = (1, 2, -1) …(vi)
Now putting μ = -2:
Direction Ratio = (2(-2) + 3, (-2) + 3, -2)
⇒ Direction Ratio = (-4 + 3, -2 + 3, -2)
⇒ Direction Ratio = (-1, 1, -2) …(vii)
Using the direction ratios in (vi) and (vii) in equation (i);
And,
Therefore, the two required lines are and
Question:12
Answer:
Given, two lines whose direction cosines are l + m + n = 0 - (i); and l² + m² - n² = 0 - (ii). We need to find the angle between these lines.
First, we must find the values of l, m and n.
From equation (i), l + m + n = 0
=> n = - l - m
=> n = -(l + m) …(iii)
If we substitute the value of n from (i) in (ii),
⇒ l = 0 or m = 0
Putting l = 0 in equation (i),
=> 0 + m + n = 0
=> m + n = 0
=> m = -n
If m = , then
n = -m = -
Hence, direction ratios (l, m, n) = (0, , -)
=> Position vector parallel to these given lines =
Now, putting m = 0 in equation (i),
=> l + 0 + n = 0
=> l + n = 0
=> l = -n
If n = , then
l = -n = -
Hence, direction ratios (l, m, n) = (-, 0, )
=> Position vector parallel to these given lines =
From the theorem, we get the angle between the two lines whose direction ratios are d_{1} and d_{2} as:
If we substitute the values of d_{1} and d_{2}, we get
Solving the numerator,
Solving the denominator,
Substituting the values in θ,
Therefore, the required angle between the lines is π/3.
Question:13
Answer:
Given: direction cosines of a variable line in two adjacent positions are l, m, n and
We have to prove that the small angle between the two positions is given by
We know, the relationships between direction cosines is given as
Also,
Let
We know, angle between two lines =
Here, the angle is very small because the line is variable in different although adjacent positions. According to the question, this small angle is
Therefore,
Substituting the values of the two vectors, we get
The dot product of 2 vectors is calculated by obtaining the sum of the product of the coefficients of
Or,
We know,
On the left-hand side, the angle is . On the right hand side, it becomes half, that is, .
Similarly replacing by in LHS, then making the angle on the RHS half,
We get:
Since is a very small angle, will be much smaller. Hence will also be very small in value.
Hence, proved.
Question:14
Answer:
We have the points O (0, 0, 0) and A (a, b, c) where a, b, and c are direction ratios. We need to find the direction cosines of line OA and the equation of the plane through A at right angle to OA.
To begin with,
We know, if (a, b, c) are the direction ratios of a given vector, then its direction cosines will be:
According to the question, the direction ratios are (a, b, c), therefore the direction cosines of the vector OA are the same as the above formula, that is,
Given, the plane is perpendicular to OA. We know, a normal is a line or vector which is perpendicular to a given object. Therefore, we can say:
Also, the vector equation of a plane where the normal is passing through the plane and passing through is,
Where
Here, the given point in the plane is A (a, b, c).
Substituting the vectors respectively, we get:
Upon simplifying this, we get:
Hence, the required equation of the plane is a² + b² + c² = ax + by + cz.
Question:15
Answer:
Given, we have 2 systems of rectangular axes. Both the systems have the same origin, and there is a plane that cuts both systems.
One system is cut at a distance of a, b, c.
The other system is cut at a distance of a’, b’, c’.
To prove:
Proof: Since a plane intersects both the systems at distances a, b, c, and a’, b’, c’ respectively, this plane will have different equations in the two different systems.
Let us consider the equation of the plane in the system with distances a, b, c to be:
Let us consider the equation of the plane in the system with distances a’, b’, c’ be:
According to the question, the plane cuts both the systems from the origin. We know, the perpendicular distance of a plane from the origin is given by:
(where not all a, b, and c are zero)
Therefore, the perpendicular distance from the origin of the first plane is:
And, the perpendicular distance from the origin of the second plane:
We also know, if two systems of lines have the same origin, their perpendicular distances from the origin to the plane in both systems are equal.
Therefore,
Cross-multiplying,
Squaring both sides,
Or
Hence, proved.
Question:16
Answer:
Given, the perpendicular from the point (let) C (2, 3, -8) to the line of which the equation is,
This can be re-written as,
Hence, the vector equation of the line is,
We must find the foot of the perpendicular from the point C (2, 3, -8) to given line, as well as the perpendicular distance from the given point C to the line.
To start with, let us locate the point of intersection between the point and the line.
Let us take,
We have,
Therefore, the coordinates of any point on the given line is
Let us consider the foot of the perpendicular from C(2, 3, -8) on line to be
Therefore, the direction ratios of
Also, the direction ratio of the line is, (-2, 6, -3).
Since L is the foot of the perpendicular on the line,
Sum of the product of these direction ratios and (-2, 6, -3) = 0.
If we substitute this value of λ in , we get
Now, we must calculate the perpendicular distance of point C from the line, that is point L.
In other words, we need to find
We know,
Substituting λ = 1,
To find
Therefore, the foot of the perpendicular from the point C to the given line is (2, 6, -2) and the perpendicular distance is units.
Question:17
Find the distance of a point (2, 4, -1) from the line
Answer:
Given, the point P (2, 4, -1), the equation of the line is
We must find the distance of point P from this line.
Note, to find the distance between a point and a line, we should get foot of the perpendicular from the point on the line.
Let, P(2, 4, -1) be the given point and be the given line.
Direction ratio of the line L is (1, 4, -9) …(i)
Let us find any point on this line.
Taking L,
Therefore, any point on the line L is
Let this point be , the foot of the perpendicular from the point P (2, 4, -1) on the line L.
Hence, the direction ratio of PQ is given by
=> Direction ratio of PQ= …(ii)
Also, we know, if two lines are perpendicular to each other, then the dot product of their direction ratios should be 0.
Here, PQ is perpendicular to L. We have, from (i) and (ii),
Direction ratio of L = (1, 4, -9)
Direction ratio of PQ =
Therefore,
Hence, the coordinate of Q, i.e. the foot of the perpendicular from the point on the given line is,
Now, to find the perpendicular distance from P to the line, that is point Q,
That is, to find
We know,
Substituting
Now, to find
Therefore, the distance from the given point to the given line = 7 units.
Question:18
Answer:
Given, point P (1, 3/2, 2)
The plane is 2x - 2y + 4z + 5 = 0
We must find the foot of the perpendicular from the point P to the equation of the given plane.
Also, we must find the distance from the point P to the plane.
Let us consider the foot of the perpendicular from point P to be Q.
Let Q be Q (x_{1} , y_{1} , z_{1})
So, the direction ratio of PQ is given by
(x_{1} - 1, y_{1} - 3/2, z_{1} - 2)
Now, let us consider the normal to the plane 2x - 2y + 4z + 5 = 0:
It is obviously parallel to PQ, since a normal is a line or vector that is perpendicular to a given object. The direction ratio simply states the number of units to move along each axis.
For any plane, ax + by + cz = d, where, a, b, and c are normal vectors to the plane.
Hence, the direction ratios are (a, b, c).
Therefore, the direction ratio of the normal = (2, -2, 4) for plane 2x - 2y + 4z + 5 = 0.
The Cartesian equation of the line PQ, where P(1, 3/2, 2) and Q (x_{1} , y_{1} , z_{1}) is:
To find any point on this line,
Any point on the line is
This point is Q.
And, it was assumed that is lies on the given plane. Substituting x_{1}, y_{1}, and z_{1} in the plane equation, we get:
2x_{1} - 2y_{1} + 4z_{1} + 5 = 0
Simplifying to find the value of
Since Q is the foot of the perpendicular from the point P,
We substitute the value of in equation (i) to get:
Then, to find
Where, P = (1, 3/2, 2) and Q = (0, 5/2, 0)
Thus, the foot of the perpendicular from the given point to the plane is (0, 5/2, 0) and the distance is units.
Question:19
Answer:
Given, a line passes through a point P (3, 0, 1) and is parallel to the planes x + 2y = 0 and 3y - z = 0.
We must find the equation of this line.
Let the position vector of point P be
Or,
Let us consider the normal to the given planes, that is, perpendicular to the normal of the plane x + 2y = 0 and 3y - z = 0
Normal to the plane x + 2y = 0 can be given as
Normal to the plane 3y - z = 0 can be given as
So, is perpendicular to both these normals.
So,
Taking the 1st row and the 1^{st} column, we multiply the 1^{st} element of the row with the difference of products of the opposite elements , excluding 1^{st} row and 1^{st} column
Here,
Now, we take the 2^{nd} column and 1^{st} row, and multiply the 2^{nd} element of the row (a??) with the difference of the product of opposite elements
Here
Finally, taking the 1^{st} row and 3^{rd} column , we multiply the 3^{rd} element of the row (a??) with the difference of the product of opposite elements excluding the 1^{s}t row and 3^{rd} column.
Here
Futher simplifying it,
Therefore, the direction ratio is (-2, 1, 3) …(iii)
We know, vector equation of any line passing through a point and parallel to a vector is where
Hence, from (i) and (ii),
Putting these vectors in the equation
We get
But we know,
Substituting this,
Thus, the required equation of the line is
Question:20
Answer:
Given, a plane passes through the points (2, 1, -1) and (-1, 3, 4) and is perpendicular to the plane x - 2y + 4z = 10.
We want to find the equation of this plane.
We know, the Cartesian equation of a plane passing through (x_{1}, y_{1}, z_{1})
with direction ratios perpendicular to a, b, c for its normal is given as:
a (x - x_{1}) +b (y - y_{1}) + c (z - z_{1}) = 0
Hence,
Let us consider the equation of the plane passing through (2, 1, -1) to be
a(x – 2) + b(y – 1) + c(z – (-1)) = 0
⇒ a(x – 2) + b(y – 1) + c(z + 1) = 0 …(i)
Since it also passes through point (-1, 3, 4) we just replace x, y, z by -1, 3, and 4 respectively.
⇒ a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0
⇒ -3a + 2b + 5c = 0 …(ii)
Since a, b, and c are direction ratios and this plane is perpendicular to the plane x - 2y + 4z = 10, we just replace x, y, and z with a, b, and c respectively (neglecting 10) and we equate this to 0.
=> a - 2b + 4c = 0 …(iii)
To solve two equations x_{1}a + y_{1}b + z_{1}c = 0 and x_{2}a + y_{2}b + z_{2}c = 0, we use the formula
Similarly, to solve for equations (ii) and (iii):
That is,
Substituting these values of a, b, and c in equation (i), we get
Therefore, the required equation of the plane is 18x + 17y + 4z = 49.
Question:21
Find the shortest distance between the lines given by and
Answer:
Given two lines,
Taking equation (i),
We know, the vector equation of a line passing through a point and parallel to a vector is where
= Position vector of the point through which line passes
= Normal to the line
Comparing this with equation (iii), we get
Now take equation (ii)
Similarly from (iv)
So, the shortest distance between two lines can be represented as:
solve
Taking 1^{st} row and 1^{st} column, we multiply the 1^{st} element of the row (a??) with the difference of the product of the opposite elements , excluding the 1^{st} row and the 1^{st} column;
Here
Now, we take the 2nd column and 1st row, and multiply the 2nd element of the row (a??) with the difference of the product of opposite elements (a?? x a?? - a?? x a??)
Here
Finally, taking the 1^{st} row and 3^{rd} column , we multiply the 3rd element of the row (a??) with the difference of the product of opposite elements (a?? x a?? - a?? x a??), excluding the 1^{st} row and 3^{rd} column.
Here
Further simplifying it.
And,
Substituting the values from (v), (vi) and (vii) in d, we get
Thus, the shortest distance between the lines is 14 units.
Question:22
Answer:
Given, a plane is perpendicular to another plane 5x + 3y + 6z + 8 = 0,and also contains line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0.
We must find the equation of this plane.
We know, the equation of a plane passing through the line of intersection of the planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 is given as,
Similarly, the equation of a plane through the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0. is given by,
Thus, the direction ratio of plane in (i) is,
Since the plane in equation (i) is perpendicular to the plane 5x + 3y + 6z + 8 = 0;
we can replace x, y, z with (1 + 2λ), (2 + λ) and (3 - λ) respectively in the plane 5x + 3y + 6z + 8 = 0 (neglecting 8) and equating to 0.
This gives us,
Substituting this value of in equation (i) we get
Thus, the required equation of the plane is 51x + 15y - 50z + 173 = 0.
Question:23
Answer:
Given, the plane ax + by = 0 is rotated about its line of intersection with z = 0 by an angle
To prove: equation of the plane in its new position is
Proof: Two planes are given, ax + by = 0 …(i) and z = 0 …(ii)
We know, the equation of the plane passing through the line of intersection of the planes (i) and (ii) is
where,
The angle between the new plane and plane (i) is given as
Since the angle between two planes is equivalent to the angle between their normals, the direction ratio of normal to ax + by = 0 or ax + by +0z = 0 is (a, b, 0).
And, the direction ratio of normal to is (a, b, λ).
Also, we know, angle between 2 normal vectors of the two given planes can be given as;
If we substitute the values of these vectors, we get
We then multiply by the numerator and denominator on the right hand side of the equation to get
Applying square on both sides,
Substituting the value of in equation (iii) to find the plane equation,
ax + by + λz = 0
Hence proved.
Question:24
Answer:
Given two planes,
Also given, the perpendicular distance of the plane from the origin = 1.
We must find the equation of this plane.
We know,
Simplifying the planes,
Also, for
The equation of a plane through the line of intersection of x + 3y - 6 = 0 and 3x - y - 4z = 0 can be given as
Also, we know, the perpendicular distance of a plane, ax + by + cz + d = 0 from the origin, let’s say P, is given by
Similarly, the perpendicular distance of the plane in equation (iii) from the origin (=1 according to the question) is:
Taking the square of both sides,
First, we subsitute in eq (iii) to find the plane equation
Now, we substitute λ= -1 in eq (iii) to find the plane equation
Therefore, the equation of the required plane is -2x + 4y + 4z – 6 = 0 and 4x – 2y – 4z – 6 = 0.
Question:25
Show that the points and are equidistant from the plane and lies on the opposite of it.
Answer:
Given two points,
Also,
Where,
Therefore,
We must show that the points A and B are equidistant from the plane
5x + 2y - 7z + 9 = 0
We also need to show that the points lie on the opposite side of the plane.
Normal of the plane is,
We know, the perpendicular distance of the position vector of a point
to the plane, p: ax + by + cz + d = 0 is given as:
Where
Thus, the perpendicular distance of the point to the plane 5x + 2y - 7z + 9 = 0 having normal is given by,
Hence, the perpendicular distance of the point to the plane 5x + 2y - 7z + 9 = 0 having normal
Therefore, |D_{1}| = |D_{2}|
However, D_{1} and D_{2} have different signs.
Therefore, the points A and B will lie on opposite sides of the plane.
Hence, we have successfully shown that the points are equidistant from the plane and lie on opposite sides of the plane.
Question:26
Answer:
Given,
And the position vectors
Therefore, the line passing through A and along AB will have the equation:
and the line passing through C and along CD will have equation
Now, PQ is a vector perpendicular to both AB and CD, such that Q lies on CD and P lies on AB. Thus, coordinates of P and Q will be of the form
Hence, the vector PQ will be given as
Now, since PQ is perpendicular to both, hence the dot products of AB.PQ and CD.PQ will be equal to 0.
AB. PQ = 0 and CD. PQ = 0
Solving (iii) and (iv), we get
Putting the value of in (i) we get,
Putting the value of in (ii) we get,
Hence, position vector of P and Q will be
Question:27
Answer:
We have given
, and
Thus, we get two cases:
=> -4m + 2m - n = 0 [from (i)]
=> n = 2m
, and
m = -2l
=> 2l + 2(-2l) - n = 0
=> 2l - 4l = n
=> n = -2l
Hence, the direction ratios of one line is proportional to -2m, m or -2m or direction ratios are (-2, 1, -2) and the direction ratios of another line is proportional to l, -2l, -2l, or direction ratios are (1, -2, -2)
Thus, the direction vectors of two lines are
Also, the angle between the two lines is given by:
Now,
Therefore, the lines have a 90^{0} angle between them.
Question:28
Answer:
Let the direction vector of the 3 mutually perpendicular lines be
Let the direction vectors associated with direction cosines be
Since the lines associated with the direction vectors a, b, and c are mutually perpendicular, we get
(Since the dot product of two perpendicular vectors is 0)
=> …(1)
Similarly,
…(2)
Finally,
=> …(3)
Now, let us consider x, y and z as the angles made by direction vectors a, b, and c respectively with p.
Then,
We know, [since the sum of squares of direction cosines of a line = 1]
[from (1) and (2)]
Then, and
=> x = y = z = 0.
Therefore, the vector p makes equal angles with the vectors a, b and c.
Question:29
Distance of the point (α, β, y) is:
A. β B. |β| C. |β| + |y| D. √(α² + y²)
Answer:
Drawing a perpendicular from (α, β, y) to the y-axis gives us a foot of perpendicular with coordinates (0, β, 0).
Also, using the distance formula, we can calculate the distance between two points as:
Thus, the required distance
(Option D)
Question:30
If the direction cosines of a line are k, k, k, then:
A. k > 0
B. 0 < k < 1
C. k = 1
D. k = 1/√3 or -1/√3
Answer:
We know that the sum of squares of the direction cosines of a line = 1
=> k² + k² + k² = 1
=> 3k² = 1
(Option D)
Question:31
The distance of the plane from the origin is:
A. 1
B. 7
C. 1/7
D. None of these
Answer:
Given plane is
Let
=> n is a unit vector
Thus, the equation of the plane is of the form , where n is
the unit vector and d is the distance from the origin.
Comparing, we get d =1, hence the distance of the plane from origin is 1
(Option A)
Question:32
The sine of the angle between the straight line and
The plane 2x - 2y + z = 5 is:
A. 10/6√5
B. 4/5√2
C. 2√3/5
D. √2/10
Answer:
The equation of the line is given as
The direction vector of this line can be represented as
Also given is the equation of the plane 2x - 2y + z = 5
The normal to this plane is,
We also know that the angle between the line with the direction
vector b and the plane with the normal vector n is,
(Option D)
Question:33
The reflection of the point (α, β, γ) in the xy- plane is:
A. (α, β, 0)
B. (0, 0, γ)
C. (-α, -β, -γ)
D. (α, β, -γ)
Answer:
The equation of the XY plane is Z = 0
Given point (α, β, γ); if we draw a perpendicular from this point in the XY plane, the coordinates of the plane will be
(α, β, 0).
Let the reflection be (x, y, z)
So,
=> x = α
=> y = β
=> z = -γ
The reflection is: (α, β, -γ). (option D)
Question:34
The area of the quadrilateral ABCD, where A (0, 4, 1), B (2, 3, -1), C (4, 5, 0) and D (2, 6, 2) is equal to:
A. 9 square units
B. 18 square units
C. 27 square units
D. 81 square units
Answer:
Given, A (0, 4, 1), B (2, 3, -1), C (4, 5, 0) and D (2, 6, 2);
Since opposite vectors of this parallelogram are equal and opposite, ABCD is a parallelogram and we know the area of a parallelogram is | AB x CD|
= 9 square units (Option A).
Question:35
The locus represented by xy + yz = 0 is:
A. A pair of perpendicular lines
B. A pair of parallel lines
C. A pair of parallel planes
D. A pair of perpendicular planes
Answer:
Given, xy + yz = 0
=> x (y + z) = 0
=> x = 0 and y + z = 0
Clearly, the above equations are the equations of planes [of the form ax + by + cz + d = 0]
Also, x = 0 has the normal vector
And y + z = 0 has the normal vector
And the dot product of these two is
= 0
Hence, the planes are perpendicular (Option D).
Question:36
The plane 2x - 3y + 6z - 11 = 0 makes an angle with the x-axis. The value of α is:
A.
B.
C.
D.
Answer:
Given, the equation of the plane is 2x - 3y + 6z - 11 = 0.
The normal to this plane is,
Also, the x-axis has the direction vector
Also, we know that the angle between the line with direction vector b and the plane having the normal vector n is:
On comparing, we find (Option C)
Question:37
Answer:
We know, the equation of a plane cutting the coordinate axes at (a, 0, 0), (0, b, 0) and (0, 0, c) is given as
In this case, a = 2, b = 3, and c = 4.
Therefore, putting these values in the equation of the plane, we get
Question:38
Fill in the blanks: The direction cosines of the vector are _______.
Answer:
If l, m, and n are the direction cosines and the direction ratios of a line are a, b, and c, then we know:
According to the question,
a = 2, b = 2, c = -1
Then
Thus, the direction cosines are
l = 2/3, b = 2/3, c = -1/3
Question:39
Fill in the blanks: The vector equation of the line is __________.
Answer:
The equation of the line is given as
Clearly, the line passes through A (5, -4, 6) and has the direction ratios 3, 7, and 2.
Also, the position vector of A is
The direction vector of the given line will be:
Also, the vector equation of a line passing through the given point whose position vector is a and b is:
Hence, the required equation of the line will be:
Question:40
Fill in the blanks: The Cartesian equation of the plane is ______.
Answer:
By expanding the dot product given in the question, we can get the Cartesian equation of the plane.
Given:
Putting
Thus, the required Cartesian equation is x + y - z = 2.
Question:41
State True or False for the given statement:
The unit vector normal to the plane is
Answer:
Given, the equation of the plane is x + 2y + 3z - 6 = 0
The normal to this plane will be The unit vector of this normal is:
Therefore, the statement is True
Question:42
Answer:
The position vector of the first point (3, 4, -7) is
And the position vector of the second point(1, -1, 6):
Also, the vector equation of a line passing through two points with position vectors a and b is given by:
Thus, the required line equation is
Question:43
Answer:
To begin with, we convert the given plane equation to intercept form:
where a, b, and c are the intercepts on x, y, and z, axes respectively.
Given,
Dividing this equation on both sides by 4,
On comparison, we get the intercepts -2, 4/3, and -4/5 respectively.
Therefore, the statement is True.
Question:44
State True or False for the given statement: The angle between the line and the plane is
Answer:
We know, the angle between the plane with normal vector n and the line with direction vector b is denoted by:
Given equation of the line is
Hence, its direction vector will be:
Given equation of the plane is
Hence, its normal vector will be:
Thus, we have:
Therefore, the given statement is False.
Question:45
State True or False for the given statement:
The angle between the planes and is
Answer:
In vector form, if we take θ as the angle between the two planes
and
Then
Now, the given planes are and
Here, and
Therefore,
The statement is False.
Question:46
State True or False for the given statement:
The line lies in the plane
Answer:
The equation of the line is given as
Any point lying on this line will satisfy the plane equation if the line itself lies in the plane. Also, any point on this line will have a position vector:
Given equation of the plane
If we put a in the above equation,
Thus, the line does not lie in the given plane.
Therefore, the given statement is False.
Question:47
State True or False for the given statement.
The vector equation of the line is
Answer:
The given equation of the line is
It is clear from the equation that this line passes through A (5, -4, 6) and has the direction ratios 3, 7 and 2.
The position vector of A is
And the direction vector of the line will be
We know, the vector equation of a line that passes through a given point with position vector a and b is given as
Hence, the required line equation will be:
Thus, the statement is True.
Question:48
Answer:
We know, the equation of a line in Cartesian form is
, where a, b and c are the direction ratios and (x_{1}, y_{1}, z_{1}) is a particular point on the line.
The given line is parallel to therefore it has 2, 1, 3 as direction ratios.
(a = 2, b = 1, c = 3)
The line passes through (5, -2, 4)
Substituting these values, we get the equation of line:
Thus, the given statement is False.
Question:49
Answer:
Let us take O as the origin, P as the foot of the perpendicular drawn from origin to the plane.
Then the position vector OP is:
The unit vector of n is:
Now, the equation of the plane with unit normal vector n and having a perpendicular drawn from the origin d is:
Therefore,
Equation of the given plane will be,
=> The given statement is True.
The main topics and sub topics covered in this chapter of are as follows:
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· NCERT Exemplar Class 12 Maths chapter 11 solutions define Dimension as the standard measure of an object's size and shape.
· We would use the concept of Vector algebra to make three-dimensional geometry organised and straightforward.
· In Class 12 Maths NCERT exemplar solutions chapter 11, we will study the direction aspects of a line joining two points, including the direction cosines and the ratios, discuss the equations of lines and planes in space, measure the shortest distance between two lines and learn more about the Cartesian form analytical and geometric representation.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | Three Dimensional Geometry |
Chapter 12 | |
Chapter 13 |
This entire chapter talks about the dimensional geometry which covered vector usage to measure and determine line, planes and angles.
Yes, for those who want a clear picture of how to solve questions in three-dimensional geometry, our NCERT exemplar Class 12 Maths solutions chapter 11 can be highly supportive.
The best way is to use these solutions as reference, while one is solving the questions for practicing.
Yes, these questions and NCERT exemplar Class 12 Maths solutions chapter 11 can be downloaded by using the webpage to PDF tool available online.
Here are some options you can explore to get admission in a good school even though admissions might be closed for many:
Waitlist: Many schools maintain waitlists after their initial application rounds close. If a student who secured a seat decides not to join, the school might reach out to students on the waitlist. So, even if the application deadline has passed, it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.
Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush. Reach out to the schools directly to see if they have any open seats in 10th grade.
Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.
Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.
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Choosing Your Stream:
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Here's why 2025 is more likely:
Looking Ahead (2025 Admissions):
Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore, with your academic background. Here's why:
High CAT Score: A score exceeding 99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes CAT scores heavily in the shortlisting process.
Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.
However, the shortlisting process is multifaceted:
Here's what you can do to strengthen your application:
Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.
Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.
Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.
hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
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