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# Test: Vector Algebra (Competition Level) - 2

## 30 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Vector Algebra (Competition Level) - 2

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This mock test of Test: Vector Algebra (Competition Level) - 2 for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Test: Vector Algebra (Competition Level) - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Vector Algebra (Competition Level) - 2 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Vector Algebra (Competition Level) - 2 exercise for a better result in the exam. You can find other Test: Vector Algebra (Competition Level) - 2 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### The differential equation of all parabola each of which has a latus rectum 4a and whose axis parallel to the x-axis is

Solution:

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

QUESTION: 2

Solution:

QUESTION: 3

### and   for positive numbers α, β and γ, then value of

Solution:

QUESTION: 4

are  any  two vectors  of magnitudes 1 and 2  respectively, and  then  the  angle  between

Solution:

QUESTION: 5

Volume of parallelopiped determined by vectors  Then the volume of the parallelopiped determined by vectors

Solution:

QUESTION: 6

4-points whose position vector  are coplanar and  then the least value of

Solution:

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

QUESTION: 7

are unit vectors, satisfying then value of

Solution:

QUESTION: 8

and are linearly dependent vectors and

Solution:

QUESTION: 9

If  are mutually perpendicular vectors and

Solution:

Taking dot product with   respectively

QUESTION: 10

If  are non-collinear unit vectors and

Solution:

QUESTION: 11

are perpendicular to respectively and if

Solution:

QUESTION: 12

is a unit vector such that  and

Solution:

QUESTION: 13

are three unit non-coplanar vectors such that is a vector such that

Solution:

QUESTION: 14

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

Solution:

Hence, equation of acute angle bisector of

y = x
or x – y = 0
Now, distance of

QUESTION: 15

Least value of the volume of the parallelopiped formed by the vectors

Solution:

QUESTION: 16

If   are non-coplanar vectors then

Solution:

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}

QUESTION: 17

Equation of the plane containing the straight line  and perpendicular to the plane containing the straight lines

Solution:

Plane containing two lines

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

QUESTION: 18

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

Solution:

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

QUESTION: 19

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

Solution:

The plane passing through the intersection line of given planes is

Its distance from the point

∴ Required equation of plane is

QUESTION: 20

Two lines   are coplanar. Then α can not  take value

Solution:

As L1, L2 are coplanar, therefore

QUESTION: 21

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

Solution:

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)

QUESTION: 22

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

Solution:

Similarly, taking dot product with vector

And, taking dot product with vector

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

QUESTION: 23

The image of line  in the plane 2x - y + z + 3 = 0 in the line

Solution:

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

QUESTION: 24

The distance of the point (1,–5,9) form the plane x - y + z = 5 measured along a straight line x = y = z is

Solution:

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

QUESTION: 25

Perpendiculars are drawn from points on the line  to the plane x + y + z = 3. The feet of perpendiculars lie on the line

Solution:

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

QUESTION: 26

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

Solution:

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

QUESTION: 27

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

Solution:

Distance from origin = 3p

QUESTION: 28

If the lines  are coplanar, then k can have

Solution:

Lines are coplanar

Exactly two values.

QUESTION: 29

If the angle θ between the line  and plane  such that   Then, value of λ is

Solution:

QUESTION: 30

The point of intersection of the lines  must be

Solution:

As the two lines intersect, we must have

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