JEE Advanced (Single Correct MCQs): Vector Algebra and Three Dimensional Geometry - JEE MCQ

 if any two vector are equal out of 

= 0

where θ is angle between 

 where θ is angle between 

Vol. of parallelopiped = 

Three pts A, B, C are collinear if 

Given that   are non coplanar

a, b, c are distinct non negative numbers and the vectors  are coplanar.

⇒ c² – ac – ab + ac = 0
⇒ c² = ab Þ  a, c, b are in G.P.
∴ c is the G.M. of  a and b.

We have 

[Internal divison]

[external divison]

Let the given position vectors be of point A, B and C respectively, then

⇒ ΔABC is an equilateral Δ.

where x² + y² + z² = 1 ....(1)

⇒ x – y = 0 ⇒ x = y ...(2)

⇒ x + y + z = 0
⇒ 2x + z = 0 (using (2))
⇒ z = – 2x ...(3) From (1), (2) and (3)

 are non-coplanar]

Substituting values of    in (1), we get

As c is coplanar with a and b, we take,

c = αa + βb  ...(1)

where α,β are scalars.
As c is perpendicular to a , c.a = 0
∴ From (1) we get, 0 = α a.a + β b.a

  (by triangle law)

Given that   are vectors such that 

P₁ is the plane determined by vectors
∴ Normal vectors  will be given by 
Similarly, P₂ is the plane determined by vectors 
∴ Normal vectors   to P₂ will be given by
Substituting the values of 

and hence the planes will also be parallel to each other.
Thus angle between the planes = 0

are unit coplanar vectors,   and   a re a lso coplan a r vector s. bein g lin ea r combination of 

= 1(1+ x - y - x + x²) -1(x²-y) = 1
∴   Depends neither on x nor on y.

Given that   are two unit vectors

Given that   and u is a unit vector 


which is max. when cos θ =1

As the line  lies in th e plan e 2x - 4 y +z= 7, the point (4, 2, k) through which line passes must also lie on the given plane and hence 2 × 4 – 4 × 2 + k = 7  ⇒ k = 7

Vol. of parallelopiped formed by

= 1(1- 0) - a(0 - a²) +1(0-a) = 1 + a³-a

⇒ x = 3 + m,y = k +2μ and z = μ
Since above lines intersect
         ...(1)
3λ - 1 = 2μ + k            ...(2)       
μ = 4λ+1                    ..(3)

Solving (1) and (3) and putting the value of  l and m in (2) we get, k = 9/2

Let the eqⁿ of variable plane be   which meets the axes at A (a, 0, 0), B (0, b, 0) and C (0. 0, c).
∴ Centroid of 

and it satisfies the relation

             ...(1)

Also given that the distance of plane  from (0, 0, 0) is 1 unit.

From (1) and (2), we get 

We observe that

 is a set of orthogonal vectors.

The equation of plane through the point (1, –2, 1) and perpendicular to the planes 2 x - 2 y +z=0 and x - y + 2z=4 is given by

It’s distance from the point (1, 2, 2) is

A vector in  the plane of   is

Projection of