The differential equation of all parabola each of which has a latus rectum 4a and whose axis parallel to the x-axis is
Let the equation of the parabola be (y – k)2 = 4a(x – h), where h and k are parameters
2 ( y - k ) dy/dx = 4a
( y - k ) dy/dx = 2a
( y - k ) d^2y/dx^2 + (dy / dx)^2 = 0
2a d^2y/dx^2 + (dy / dx)^3 = 0
Order : 2
Degree : 3
are three vectors of magnitudes
and
for positive numbers α, β and γ, then value of
are any two vectors of magnitudes 1 and 2 respectively, and
then the angle between
Volume of parallelopiped determined by vectors Then the volume of the parallelopiped determined by vectors
4-points whose position vector are coplanar and
then the least value of
As the vectors are complainer sum of coefficient =0
sinα + 2sinβ + 3sinγ = 1
this can be also LHS can also be called to be dot product of two vectors (1,2,3) and (sinα,sinβ,sinγ)
dot product of these two vectors is
sinα + 2sinβ + 3sinγ
aˉ.bˉ=|a||b|cosθ
(sinα+2sinβ+3sinγ)/1 = (sin2α+sin2β+sin2γ)×14cosθ
1/14cosθ = sin^2α+sin^2β+sin^2γ
∴ Minimum value is 1/14
are unit vectors, satisfying
then value of
and
are linearly dependent vectors and
If are mutually perpendicular vectors and
Taking dot product with respectively
If are non-collinear unit vectors and
are perpendicular to
respectively and if
is a unit vector such that
and
are three unit non-coplanar vectors such that
is a vector such that
In be the position vectors of X, Y and Z with respect of the origin O, respectively. If the distance of Z from the bisector of the acute angle of
then sum of all possible values of β is
Hence, equation of acute angle bisector of
y = x
or x – y = 0
Now, distance of
Least value of the volume of the parallelopiped formed by the vectors
If are non-coplanar vectors then
A) {3} is a set containing only one element i.e. 3, This set does not belong to {1, 3, 5}
B) {1} is a set containing only one element i.e. 1, This set does not belong to {1, 3, 5}
C) {3, 5} is a set containing two elements i.e. 3 & 5, This set does not belong to {1, 3, 5}
D) 3 is an element of the set {1, 3, 5}. Therefore, 3 ϵ {1, 3, 5}
Equation of the plane containing the straight line and perpendicular to the plane containing the straight lines
Plane containing two lines
Now equation of plane containing the line and perpendicular to the plane
If the distance of the point P (1, –2, 1) from the plane x + 2y – 2z = α , where α > 0, is 5, then the foot of the perpendicular from P to the plane is
As perpendicular distance of x + 2y – 2z = α from the point (1, –2, 1) is 5
∴ Plane becomes x + 2y – 2z – 10 = 0
It lies on x + 2y – 2z – 10 = 0
The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3 and at a distance from the point (3, 1, –1) is
The plane passing through the intersection line of given planes is
Its distance from the point
∴ Required equation of plane is
Two lines are coplanar. Then α can not take value
As L1, L2 are coplanar, therefore
A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the points of intersection are given by
Let a point on the line x = y + a = z is
and a point on the line
then direction ratio of the line joining these points are
If it represents the required line, then
on solving we get λ = 3a, μ = 2a
∴ The required points of intersection are
Or (3a, 2a, 3a) and (a, a, a)
be three non-coplanar unit vectors such that the angle between every pair of them is
where p, q and r are scalars, then the value of
Similarly, taking dot product with vector
And, taking dot product with vector
Solving, (1), (2) and (3), we get
The image of line in the plane 2x - y + z + 3 = 0 in the line
3(2) + 1(-1) + (-5) (1) = 0
Given line and given plane are parallel
∴ Image line is also plane are parallel
Image of A(1,3,4) w.r.t o given plane lies on the image line.
Equation of the normal to the plane is
Any point on the line B = (2r + 1, -r + 3, r + 4) If B is the image of A(1,3,4) then mid point of AB lies on the plane.
Mid point lies in the given plane
The distance of the point (1,–5,9) form the plane x - y + z = 5 measured along a straight line x = y = z is
Let P=(1,–5,9)
Let Q be a point on the given plane such that PQ is parallel to given line The equation of the line PQ is
Let Q = (1 + t, t - 5, t + 9)
Sub Q in the given plane, t =-10
Perpendiculars are drawn from points on the line to the plane x + y + z = 3. The feet of perpendiculars lie on the line
Any point on line
This point lies on the plane x + y + z = 3.
So, 4t = 6
∴ t = 3/2
Thus, point B is (1, –5/2, 9/2).
Also, equation of line AC is
So, any point on this line is C(s – 2, s – 1, s).
This point lies on the plane x + y + z = 3.
∴ s = 2
Thus, point C is (0, 1, 2).
Therefore, directions ratios of BC are (1, –7/2, 5/2) or (2, –7, 5).
Hence, feet of perpendiculars lies on
A variable plane at a distance of 1 unit from the origin cut the co-ordinate axis at A, B and C. If the centroid D (x, y, z) of DABC satisfy the relation then value of k is
Equation of plane in intercept form is
Centroid of ΔABC(a/3, b/3, c/3)
It satisfies the relation
∴ 9/a2 + 9/b2 + 9/c2 = k
Perpendicular distance is 1 unit
A variable plane is at a constant distance 3p form the origin and meets the axes in A, B and C. The locus of the centroid of the triangle ABC is
Distance from origin = 3p
If the lines are coplanar, then k can have
Lines are coplanar
Exactly two values.
If the angle θ between the line and plane
such that
Then, value of λ is
The point of intersection of the lines must be
As the two lines intersect, we must have
∴Point of intersection is (a + a ', b + b ', c + c ')
⇒ (B) is correct
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