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A plane which passes through the point (3, 2, 0) and the line
As the point (3, 2, 0) lies on the given line
∴ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points (3, 2, 0) and (4, 7, 4)
∴ x – y + z = 1 is the required plane.
and the angle between then is equal to
are vectors such that then
are vectors show that and
thus what will be the value of given that
If the vectors are such that form a right handed system then
Since form a right handed system,
and are two vectors and a vector such that
Above three conditions will be satisfied for nonzero vectors if and only if
The d.r. of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle π /4 with plane x + y = 3 are
Equation of plane through (1, 0, 0) is a (x – 1) + by + cz = 0 ...(i)
(i) passes through (0, 1, 0). –a + b = 0 ⇒ b = a; Also, cos 45°
So d.r of normal are a, a
Let is a unit vector such that is equal to
since perpendicular
A particle acted on by constant forces and is displaced from the point to the point The total work done by the forces is
The vectors are the sides of a triangle ABC. The length of the median through A is
The shortest distance from the plane 12x + 4 y + 3z = 327 to the sphere x^{2} + y^{2} + z^{2} + 4x  2y  6z = 155 is
Shortest distance = perpendicular distance between the plane and sphere = distance of plane from centre of sphere – radius
The two lines x = ay + b , z = cy + d and x = a`y + b`, z = c`y + d` will be perpendicular, if and only if
For perpendicularity of lines aa'+1 + cc' = 0
The lines arecoplanar if
are 3 vectors, such that , then is equal to
The radius of the circle in which the sphere
x^{2} + y^{2} + z^{2} + 2x  2y  4z  19 = 0 is cut by the plane x + 2y + 2z + 7 = 0 is
Centre of sphere = (–1, 1, 2) Radius of sphere
Perpendicular distance from centre to the plane
A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1) B(2, 1, 3) and C(–1, 1, 2). Then the angle between the faces OAB and ABC will be
Vector perpendicular to the face OAB
Vector perpendicular to the face ABC
Angle between the faces = angle between their normals
and vectrs (1, a,a^{2}),(1, b, b^{2}) and (1, c,c^{2}) are non coplanar, then the product abc equals
Consider points A , B , C and D with position vectors and respectively. Then ABCD is a
∴ None of the options is satisfied
If are three non  coplanar vectors, then equals
Two system of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a' , b' , c' from the origin then
Eq. of planes be
(⊥ r distance on plane from origin is same.)
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y +4z + 5 = 0 is
The planes ar e 2x + y + 2z  8=0 . ...(1)
and 4x + 2y + 4z + 5 = 0
...(2)
∴ Distance between (1) and (2)
A line with direction cosin es proportional to 2, 1, 2 meets each of the lines x = y + a = z an d x + a = 2y = 2z . The coordinates of each of the points of intersection are given by
Let a point on th e lin e x = y + a = z is
(λ, λ  a,λ) and a point on th e lin e x + a = 2 y = 2z
then direction ratio of the line joining these poin ts are
If it respresents the required line, then
on solving we get λ = 3a, μ = 2a
∴ The required points of intersection are (3a, 3aa,3a) and
or (3a, 2a, 3a) and (a,a,a)
If the straight lines
with parameters s and t respectively, are coplanar, then λ equals.
The given lines are
The lines are coplanar, if
The intersection of the spheres
x^{2} + y^{2} + z^{2} + 7x  2y  z = 13 and x^{2} + y^{2} + z^{2} 3x + 3y + 4z = 8
is the same as the intersection of one of the sphere and the plane
The equation s of spheres are
S_{1} : x^{2} + y^{2} + z^{2} + 7x  2y  z 13 = 0 and
S_{2} : x^{2} + y^{2} + z^{2 } 3x + 3y + 4z8 = 0
Their plane of intersection is
S_{1}  S_{2} = 0 ⇒ 10x  5y  5z  5 = 0
⇒ 2x yz = 1
be three non zero vectors such that no two of these are collinear. If the vector is collinear with is collinear with (λ being some nonzero scalar) then equals
Let wh ere t and s are scalars. Adding, we get
A particles is acted upon by constant forces and which displace it from a point to the point The work done in standard units by the forces is given by
If are noncoplanar vectors and l is a real number, then the vectors are non coplanar for
coplanar if
∵ Forces are noncoplanar for all λ, except
be such that If the projection is equal to that of and are perpendicular to each other then
be nonzero vectors such that is the acute angle between the vectors then sinθ equals
If C is the mid point of AB and P is any point outside AB, then
If the angel θ between the line and the plane is such that sin θ = 1/3 then the value of λ is
If θ is the angle between line and plane then is the angle between line and normal to plane given by
The angle between the lines 2x = 3y = – z and 6x = – y = – 4z is
The given lines are 2x = 3y = z
[Dividing by 6]
and 6 x =  y = 4z
[Dividing by 12]
∴ Angle between two lines is
If the plane 2ax – 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres x^{2} + y^{2} + z^{2} + 6x  8y 2z = 13 and x^{2} + y^{2} + z^{2}  10 x + 4y  2z = 8 then a equals
Centers of given spheres are (– 3, 4, 1) and (5, – 2, 1).
Mid point of centres is (1, 1, 1).
Satisfying this in the equation of plane, we get 2a  3a + 4a + 6 = 0 ⇒ a =2.
The distance between the line and the plane
A point on line is (2, –2, 3) its perpendicular distance from the plane x + 5 y +z – 5 = 0 is
For any vector the value of is equal to
If non zero numbers a, b, c are in H.P., then the straight line always passes through a fixed point. That point is
Let a, b and c be distinct non negative numbers. If the vectors lie in a plane, then c is
Vector are coplanar
If are non coplanar vectors and λ is a real number then
Let and depends on
Hence is independent of x and y both.
The plane x + 2y – z = 4 cuts the sphere x^{2} + y^{2 }+ z^{2} – x + z – 2 = 0 in a circle of radius
Perpendicular distance of centre from x + 2y – 2 = 4 is given by
radius of sphere
where are any three vectors such that are
The values of a, for which points A, B, C with position vectors respectively are the vertices of a right angled triangle with C = π/2 are
The two lines x = ay+b , z = cy+d ; and x = a' y+b' , z = c' y+d' are perpendicular to each other if
Equation of lines
Line are perpendicular ⇒ aa '+ 1 + cc '= 0
The image of the point (–1, 3, 4) in the plane x  2y = 0 is
N is mid point of PP'
If a line makes an angle of π /4 with the positive directions of each of x axis and y axis, then the angle that the line makes with the positive direction of the zaxis is
Let the angle of line makes with the positive direction of zaxis is a direction cosines of line with the +ve directions of xaxis, yaxis, and zaxis is l, m, n respectively.
Hence, angle with positive direction of the
If are unit vectors and θ is the acute angle between them, then is a unit vector for
Given is acute angle between and
Hence, there is exactly one value of θ for which is a unit vector..
If (2, 3, 5) is one end of a diameter of the sphere x^{2} + y^{2} + z^{2} – 6x – 12y – 2z + 20 = 0, then the cooordinates of the other end of the diameter are
For given sphere centre is (3, 6, 1) Coordinates of one end of diameter of the sphere are (2, 3, 5). Let the coordinates of the other end of diameter are (α, β,γ)
∴ Coordinate of other end of diameter are (4, 9, –3)
Let If the vectors in the plane of then x equals
lies in the plane of
Let L be the line of intersection of the planes 2x + 3y + z = landx + 3y + 2z= 2. IfL makes an angle α with the positive xaxis, then cos α equals
Let the direction cosines of line L be l, m, n, then 2l + 3m + n = 0 ....(i)
and l + 3m + 2n = 0 ....(ii)
on solving equation (i) and (ii), we get
Line L, makes an angle α with +ve xaxis
The vector lies in the plane of the vectors and bisects the angle between Then which one of the following gives possible values of α and β?
The nonzero vectors are are related by and Then the angle between
and are opposite in direction
∴ Angle between
The line passing through the points (5, 1, a) and (3, b, 1) crosses the yzplane at the point Then
Equation of line through (5, 1, a) and
∴ Any point on this line is a
If the straight lines and intersect at a point, then the integer k is equal to
The two lines intersect if shortest distance between them is zero i.e.
∵ k is an integer, therefore k = –5
Let the line lie in the plane x + 3y – αz + β = 0. Then (α, β) equals
x + 3y – αz + β = 0
∴ Pt (2, 1, – 2) lies on the plane
i.e. 2 + 3 + 2α + β = 0 or 2α + β + 5 = 0 ....(i)
Also normal to plane will be perpendicular to line,
∴ 3 × 1 – 5 × 3 + 2 × (– α) = 0 ⇒ α = – 6
From equation (i) then, b = 7
∴ (α, β) = (– 6, 7)
The projections of a vector on the three coordinate axis are 6, –3, 2 respectively. The direction cosines of the vector are :
Let P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) be the initial and final points of the vector whose projections on the three coordiante axes are 6, – 3, 2 then
So that direction ratios of
∴ Direction cosines of are
If are noncoplanar vectors and p, q are real numbers, then the equality holds for :
∴ Exactly one value of (p, q)
Let Then the vector satisfying
which is not possible
If the vectors and are mutually orthogonal, then (λ, μ)=
Since are mutually orthogonal
On solving (i) and (ii), we get λ = 3, μ = 2
Statement 1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5.
Statement 2: The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4).
A(3, 1, 6); B = (1, 3, 4) Midpoint of AB = (2, 2, 5) lies on the plane. and d.r’s of AB = (2, –2, 2)
d.r’s of normal to plane = (1, –1, 1).
Direction ratio of AB and normal to the plane are proportional therefore, AB is perpendicular to the plane
∴ A is image of B
Statement2 is correct but it is not correct explanation.
A line AB in threedimensional space makes angles 45° and 120° with the positive xaxis and the positive yaxis respectively. If AB makes an acute angle θ with the positive zaxis, then θ equals
Direction cosines of the line :
where θ is the angle, which line makes with positive zaxis.
If the angle between the line and the plane then λ equals
If θ be the angle between the given line and plane, then
then the value
The vectors are not perpendicular and are two vectors satisfying Then the vector
Statement1: The point A(1, 0, 7)) is the mirror image of the point B(1, 6, 3) in the line :
Statement2: The line bisects the line segment joining A(1, 0, 7) and B(1, 6, 3) .
The direction ratios of the line segment joining points
A(1, 0, 7) and B(1, 6, 3) are 0, 6, – 4.
The direction ratios of the given line are 1, 2, 3.
Clearly 1 × 0 + 2 × 6 + 3 × (4) = 0
So, the given line is perpendicular to line AB.
Also , the mid point of A and B is (1, 3, 5) which lies on the given line.
So, the image of B in the given line is A, because the given line is the perpendicular bisector of line segment joining points A and B, But statement2 is not a correct explanation for statement1.
be two unit vectors . If the vectors are perpendicular to each other,, then the angle between
are perpendicular to each other
A equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is :
Given equation of a plane is x – 2y + 2z – 5 = 0
So, Equation of parallel plane is given by x – 2y + 2z + d = 0
Now, it is given that distance from origin to the parallel plane is 1.
So equation of required plane x – 2y + 2z ± 3 = 0
If the line intersect, then k is equal to:
Given lines in vector form are
These will intersect if shortest distance between them = 0
Let ABCD be a parallelogram such that and ∠BAD be an acute angle. is the vector that coincide with the altitude directed from the vertex B to the side AD, then
Let ABCD be a parallelogram such that
be an acute angle.
We have
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
2x + y + 2z – 8 = 0 …(Plane 1)
…(Plane 2)
Distance between Plane 1 and 2
If the lines are coplanar, then k can have
Given lines will be coplanar
If the vectors are the sides of a triangle ABC, then the length of the median through A is
∵ M is mid point of BC
Length of median AM
The image of the line in the plane 2x  y +z + 3 = 0 is the line:
The angle between the lines whose direction cosines satisfy the equations l + m +n= 0 and l^{2} = m^{2} + n^{2} is
Given l + m + n = 0 and l^{2} = m^{2} + n^{2}
Now, (–m –n)^{2} = m^{2} + n^{2}
⇒ mn = 0 ⇒ m = 0 or n = 0
If m = 0 then l = –n
then λ is equal to
Let be three nonzero vectors such that no two of them are collinear and If θ is the
angle between vectors then a value of sin θ is :
are non collinear, the above equation is possible only when
The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is :
Equation of the plane containing the lines
2x – 5y + z = 3 and x + y + 4z = 5 is
2x – 5y + z – 3 + λ (x + y + 4z – 5) = 0
⇒ (2 + λ) x + (–5 + λ) y + (1 + 4λ)z + (–3 – 5λ) = 0 ...(i)
Since the plane (i) parallel to the given plane x + 3y + 6z = 1
Hence equation of the required plane is
The distance of the point (1, 0, 2) from the point of intersection of the line and the plane
General point on given line ≡ P(3r + 2, 4r – 1, 12r + 2)
Point P must satisfy equation of plane
(3r + 2) – (4r – 1) + (12r + 2) = 16
11r + 5 = 16
r = 1
P(3 × 1 + 2, 4 × 1 – 1, 12 × 1 + 2) = P(5, 3, 14)
distance between P and (1, 0, 2)
If the line, lies in the plane, lx + my – z = 9,then l^{2} + m^{2} is equal to :
Line lies in the plane ⇒ (3, –2, –4) lie in the plane
⇒ 3ℓ – 2m + 4 = 9 or 3ℓ – 2m = 5 ..... (1)
Also, ℓ, m,–1 are dr's of line perpendicular to plane and 2, –1, 3 are dr's of line lying in the plane
⇒ 2ℓ – m – 3 = 0 or 2ℓ – m = 3 .....(2)
Solving (1) and (2) we get ℓ = 1 and m = –1
⇒ ℓ^{2} + m^{2} = 2
be three unit vectors such that is not parallel to then
the angle between is:
On comparing both sides
where θ is the angle between
The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :
Putting these in eq^{n} of plane :
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