JEE Advanced Level Test: Vector Algebra- 1 - JEE MCQ

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

 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

Taking dot product with   respectively

Hence, equation of acute angle bisector of

y = x
or x – y = 0
Now, distance of 

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}

Plane containing two lines 

Now equation of plane containing the line  and perpendicular to the plane 

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 plane passing through the intersection line of given planes is

Its distance from the point  


∴ Required equation of plane is

As L₁, L₂ are coplanar, therefore

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)

Similarly, taking dot product with vector 

 And, taking dot product with vector 

Solving, (1), (2) and (3), we get

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 

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

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 

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

Distance from origin = 3p 

Lines are coplanar

Exactly two values.

As the two lines intersect, we must have


∴Point of intersection is (a + a ', b + b ', c + c ')
⇒ (B) is correct