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# Competition Level Test: Vector

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

Competition Level Test: Vector for JEE 2023 is part of Mathematics (Maths) Class 12 preparation. The Competition Level Test: Vector questions and answers have been prepared according to the JEE exam syllabus.The Competition Level Test: Vector MCQs are made for JEE 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Competition Level Test: Vector below.
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Competition Level Test: Vector - Question 1

### If the vector  is collinear with the vector (2√2, -14) and  = 10, then

Detailed Solution for Competition Level Test: Vector - Question 1

a = 2(2)½ i -  j + 4k
|b| = 10
[(2(2λ)2 + (1λ)2 + (4λ)2]½
=[ 8λ2 + λ2 + 16λ2]½ = 10
= 25λ2 = (10)2
25λ2 = 100
λ = +-2
b = +-(4(2)½ i - 2j + 8k)
Therefore, 2a +- b

Competition Level Test: Vector - Question 2

### The vertices of a triangle are A(1, 1, 2), B(4, 3, 1) and C(2, 3, 5). A vector representing the internal bisector of the angle A is

Detailed Solution for Competition Level Test: Vector - Question 2

Let AD is the bisector of ∠A.Then,
BD/DC = AB/ACeq(1)
Given the vertices of triangle,
AB = √32 + 22 + 12 = √14
AC = √12 + 22 + 32 = √14
As AB = BC, So, BD = DC(from eq(1))
It means D is middle point of BC. So, vertices of D will be (3,3,3).
So, vector AD will be 2iˆ+ 2jˆ+ kˆ.

Competition Level Test: Vector - Question 3

### Let  . The point of intersection of lines

Competition Level Test: Vector - Question 4

If  and   then  is equal to

Competition Level Test: Vector - Question 5

Angle between diagonals of a parallelogram whose side are represented by

Detailed Solution for Competition Level Test: Vector - Question 5

D1 = a+b
D2 = a-b
D1 = 3i + 0j + 0k
D2 = i + 2j + 2k
|D1| = 3
D1.D2 = |D1| . |D2| . cos θ
3 + 0 + 0 = (3) . (3) cos θ
3 = 9 cosθ
cos-1 = (⅓)

Competition Level Test: Vector - Question 6

Vector  make an angle θ = 2π/3. if  , then is equal to

Detailed Solution for Competition Level Test: Vector - Question 6

Competition Level Test: Vector - Question 7

Unit vector perpendicular to the plane of the triangle ABC with position vectors  of the vertices
A, B, C is

Detailed Solution for Competition Level Test: Vector - Question 7

Δ = 1/2(a * b) = 1/2(b*c) = 1/2(c*a)
2Δ = a*b = b*c = c*a
unit vector = 1/2Δ[a*b + b*c + c*a]

Competition Level Test: Vector - Question 8

The value of  is equal to the box product

Detailed Solution for Competition Level Test: Vector - Question 8

Matrix {(1,2,-1) (1,-1,0) (1,-1,-1)} [a b c]
= {1(1) -2(-1) -1(0)} [a b c]
= 3[a b c]

Competition Level Test: Vector - Question 9

If are two non-collinear vectors such that , then  is equal to

Competition Level Test: Vector - Question 10

Vector of length 3 unit which is perpendicular to  and lies in the plane of  and

Detailed Solution for Competition Level Test: Vector - Question 10

Competition Level Test: Vector - Question 11

Vector  satisfying the relation

Detailed Solution for Competition Level Test: Vector - Question 11

A * X = B
(A * X)*A = B * A
=> -A(x.A) + x(A.A) = B * A
=> -Ac + x|A|2 = B * A
x|A|2 = B * A + Ac
x = [B * A + Ac]/|A|2

Competition Level Test: Vector - Question 12

If a ,b,c are linearly independent vectors, then which one of the following set of vectors is linearly dependent ?

Detailed Solution for Competition Level Test: Vector - Question 12

xa + yb + zc = 0
We have to prove x = y = z = 0
a,b,c are non planner
x(a-b) +y(b-c) +c(c-a) = 0
Let x = 1, y = 1, z = 1
So, we get a - b + b - c + c - a = 0

Competition Level Test: Vector - Question 13

Let be vectors of length 3,4,5 respectively. Let be perpendicular to , and . then

Detailed Solution for Competition Level Test: Vector - Question 13

|A| = 3, |B| = 4, |C| = 5
Since A.(B + C) = B.(C+A) = C(A+B) = 0...........(1)
|A+B+C|2 = |A| + |B|2 + |C|2 + 2(A.B + B+C + C.A)
= 9+16+25+0
from eq(1) {A.B + B+C + C.A = 0}
therefore, |A+B+C|2 = 50
=> |A+B+C| = 5(2)1/2

Competition Level Test: Vector - Question 14

Given the vertices A (2, 3, 1), B(4, 1, –2), C(6, 3, 7) & D(–5, –4, 8) of a tetrahedron. The length of the altitude drawn from the vertex D is

Competition Level Test: Vector - Question 15

for a non zero vector  If the equations   hold simultaneously, then

Competition Level Test: Vector - Question 16

The volume of the parallelopiped constructed on the diagonals of the faces of the given rectangular parallelopiped is m times the volume of the given parallelopiped. Then m is equal to

Detailed Solution for Competition Level Test: Vector - Question 16

Vi=[a→,b→,c→]
(a→ + b→)(b→ + c→)(a→ + c→)
Vf=[(a→ + b→)(b→ + c→)(a→ + c→)]
=(a→ + b→)⋅[(b→ + c→)⋅(a→ + c→)]
=(a→ +b→)[b→⋅a→ + c→⋅a→ + b→⋅c→ + c→⋅c→]
=[b→ c→ a→]+[a→ b→ c→]
Vf=2[a→ b→ c→]
Vf=2Vi.

Competition Level Test: Vector - Question 17

If u and v are unit vectors and θ is the acute angle between them, then 2u × 3v is a unit vector for

Competition Level Test: Vector - Question 18

The value of a, for which the points A,B,C with position vectors  and  respectively are the vertices of a right angled triangle with C = π/2 are

Detailed Solution for Competition Level Test: Vector - Question 18

Competition Level Test: Vector - Question 19

The distance between the line  and the plane  is

Competition Level Test: Vector - Question 20

A particle is acted upon by constant forces which displace ot from a point   to the point . The workdone in standard units by the force is given by

Competition Level Test: Vector - Question 21

If  are non-coplaner vectors and λ is a real number, then the vectors  are non-coplaner for

Competition Level Test: Vector - Question 22

Let  be non zero vectors such that , If θ is the acute angle between the vectors , then sin θ equals is

Detailed Solution for Competition Level Test: Vector - Question 22

(a→*b→)*c→ = (1/3)|b→||c→||a→|
⇒ − c→*(a→*b→) = (1/3)|b→||c→||a→|
⇒(c→.a→)b→ −(c→.b→)a→ =(1/3)|b→||c→|a→|
Now, as all of them are non-collinear, (c→⋅a→) can be 0 that means,
(c→.b→) = (−1/3)|b→||c→|
⇒|b→|.|c→|cosθ = (−1/3)|b→||c→|
⇒ cosθ = −1/3
sinθ = √1−(1/3)^2
= √8/9
= (2√2)/√3

Competition Level Test: Vector - Question 23

are three vectors, such that  then is equal to

Competition Level Test: Vector - Question 24

If  are three non-coplaner vectors, then  equals

Competition Level Test: Vector - Question 25

Competition Level Test: Vector - Question 26

The vectors  are the sides of a triangle ABC. The length of the median through A is

Detailed Solution for Competition Level Test: Vector - Question 26

The length of median through A = vector(AB + BC)/2
= (3i + 4k + 5i - 2j + 4k)/2
= (8i - 2j + 8k)/2
= 4i - j + 4k
Length  = √(16 + 1 + 16) = √33

Competition Level Test: Vector - Question 27

Competition Level Test: Vector - Question 28

If is equal to

Detailed Solution for Competition Level Test: Vector - Question 28

∣u × v∣=∣(a − b) × (a + b)∣
=2∣a × b∣      (∵a × a = b × b = 0)
and ∣a × b∣2 + (a ⋅ b)2
=(ab sinθ)+ (abcosθ)2
=a2b2
⇒∣a×b∣ = a2b2−(a ⋅ b)2
So, ∣u × v∣ = 2∣a × b∣
=2[a ^ 2b− (a ⋅ b)2]1/2
=2[(2)2(2)2−(a⋅b)2]1/2
=2(16−(a⋅b)2)1/2
∴∣a∣=∣b∣=2

Competition Level Test: Vector - Question 29

Let  and  be three non-zero vectors such that  is a unit vector perpendicular to both . if the angle between  is π/6, then is equal to

Detailed Solution for Competition Level Test: Vector - Question 29

According to the given conditions,
(c1)2+c(2)2+(c3)22 =1,a⋅c=0,b⋅c=0
and cosπ/6 = [(3)1/2]/2
​(a1b1+a2b2+a3b3]/[(a1)2+(a2)2 + (a3)^2)]^1/2 [(b1)^2 + (b2)2 + (b3)2]1/2
Thus a1c1+a2c2+a3c3=0, b1c1+b2c2+b3c3=0
and [(3)^1/2]/2[(a1)2+(a2)2 + (a3)2)1/2 ((b1)2 + (b2)2 +(b3)2]1/2
=a1b1+a2b2 +a3b3
[(a1)2+(a2)2 + (a3)2] -(a1b1 + a2b2 + a3b3)2
[(a1)2+(a2)2 + (a3)2] [(b1)2+(b2)2 + (b3)2]
= -3/4 [(a1)2+(a2)2 + (a3)2] [(b1)2+(b2)2 + (b3)2]
= 1/4 [(a1)2+(a2)2 + (a3)2] [(b1)2+(b2)2 + (b3)2]

Competition Level Test: Vector - Question 30

A point taken on each median of a triangle divides the median in the ratio 1 : 3, reckoning from the vertex.
Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is

Detailed Solution for Competition Level Test: Vector - Question 30

Let A (0,0); B (4m , 0) and C(4p , 4q)
M1 (2m + 2p, 2q)
M2 (2p , 2q) and M3 (2m , 0)
Let E , F and G be the point on the median.
E = (2m + 2p) / 4, 2q / 4) = ((m + p) / 2, q / 2)
F = ((2p + 12m) / 4, (2q + 0) / 4) = ((p + 6m) / 2, q / 2)
G = ((2m + 12p) / 4, (0 + 12 q) / 4) = ((m + 6p ) / 2, 3q)
Area of traingle ABC = 1/2
Area of traingle ABC = 1/2 {(0,0,0) (4m,0,1) (4p,4q,1)}
=1/2(16) = 8 unit
Area of triangle EFG = 1/2 {((m + p) / 2, q / 2, 1)) ((p + 6m) / 2, q / 2, 1) ((m + 6p)/2, 3q, 1)}
= 25/8 unit
ar (EFG) /ar (ABC) = {25 / 8} / 8
= 25/64

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## Mathematics (Maths) Class 12

209 videos|218 docs|139 tests