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QUESTION: 1

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

Solution:

a = 2(2)^{½ }i - j + 4k

|b| = 10

[(2(2λ)^{2} + (1λ)^{2} + (4λ)^{2}]^{½}

=[ 8λ^{2} + λ^{2} + 16λ^{2}]^{½} = 10

= 25λ^{2} = (10)^{2}

25λ^{2} = 100

λ = +-2

b = +-(4(2)^{½} i - 2j + 8k)

Therefore, 2a +- b

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

Solution:

Let AD is the bisector of ∠A.Then,

BD/DC = AB/ACeq(1)

Given the vertices of triangle,

AB = √3^{2} + 2^{2} + 1^{2} = √14

AC = √1^{2} + 2^{2} + 3^{2} = √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.

QUESTION: 3

Let . The point of intersection of lines

Solution:

QUESTION: 4

If and then is equal to

Solution:

QUESTION: 5

Angle between diagonals of a parallelogram whose side are represented by

Solution:

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)

QUESTION: 6

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

Solution:

QUESTION: 7

Unit vector perpendicular to the plane of the triangle ABC with position vectors of the vertices

A, B, C is

Solution:

**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]

QUESTION: 8

The value of is equal to the box product

Solution:

**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]

QUESTION: 9

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

Solution:

QUESTION: 10

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

Solution:

QUESTION: 11

Vector satisfying the relation

Solution:

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}

QUESTION: 12

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

Solution:

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

QUESTION: 13

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

Solution:

|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}

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

Solution:

QUESTION: 15

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

Solution:

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

Solution:

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.

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

Solution:

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

Solution:

QUESTION: 19

The distance between the line and the plane is

Solution:

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

Solution:

QUESTION: 21

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

Solution:

QUESTION: 22

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

Solution:

(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

QUESTION: 23

are three vectors, such that then is equal to

Solution:

QUESTION: 24

If are three non-coplaner vectors, then equals

Solution:

QUESTION: 25

Solution:

QUESTION: 26

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

Solution:

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

QUESTION: 27

Solution:

QUESTION: 28

If is equal to

Solution:

∣u × v∣=∣(a − b) × (a + b)∣

=2∣a × b∣ (∵a × a = b × b = 0)

and ∣a × b∣^{2} + (a ⋅ b)^{2}

=(ab sinθ)^{2 }+ (abcosθ)^{2}

=a^{2}b^{2}

⇒∣a×b∣ = a^{2}b^{2}−(a ⋅ b)^{2}

So, ∣u × v∣ = 2∣a × b∣

=2[a ^ 2b^{2 }− (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

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

Solution:

According to the given conditions,

(c1)^{2}+c(2)^{2}+(c3)^{2}2 =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}]

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

Solution:

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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