Scalar Or Dot Products

Given Vectors:

.

2

.

Now

λ × 4 + 2 × (-9) + 1 × 2 = 0

λ × 4 = 16

λ = 4

Now

λ × 5 + 2 × (-9) + 1 × 2 = 0

λ × 5 = 16

Now

2 × 3 + 3 × 2 + 4 × (-λ) = 0

- 4 λ = - 12

λ = 3

Now

λ × 1 + 3 × (-1) + 2 × 3 = 0

λ - 3 + 6 = 0

λ = -3

Given Data:

Calculation:

Given data:

Now

Consider

Given Data:

i.

Calculation:

Angle with x-axis

Angle with y-axis

Angle with z-axis

Given Data:

Vectors:

Their Dot products are 0, 5 and 8.

Calculation:

Let the required vector be,

Now,

x + y -3z =0 …Eq. 1

Similarly

x + 3y - 2z = 5 …Eq. 2

2x + y + 4z = 8 …Eq. 3

Subtract Eq. 1 from Eq. 2

(x + 3y - 2z) – (x + y -3z) = 5 - 0

⇒ 2y + z = 5 …Eq. 4

Subtract Eq. 3 from (2 × Eq. 2)

2(x + 3y - 2z) - 2x + y + 4z = (2 × 5) - 8

5y – 8z = 2 …Eq. 5

Adding Eq. 5 with (8 × Eq. 4)

8(2y + z) + (5y – 8z) = 8 × 5 + 2

⇒ 21y = 42

⇒ y = 2

From Eq. 5,

5×2 - 8z = 2

⇒ z = 1

From Eq. 1

x + y -3z =0

⇒ x + 2 – 3×1 = 0

⇒ x = 1

Vectors:

Their Dot products are 4, 0 and 2.

Calculation:

Let the required vector be,

Now,

x - y + z = 4 …Eq. 1

Similarly

2x + y - 3z = 0 …Eq. 2

x + y + z = 2 …Eq. 3

Subtract Eq. 1 from Eq. 3

(x + y + z) – (x - y + z) = 2 - 4

⇒ 2y = -2

y = - 1

Now putting the value of y in equation(2) and equation (3) we get,

2 x – 3 z = 1 …(Eq(4))

x + z = 3 ……(Eq(5))

Eq(4) – 2 × Eq (5)

- 5 z = - 5

z = 1

Now putting value of z in equation (1) we get,

x – y + z = 4

x + 1 + 1 = 4

x = 2

So the vector is,

Given Data: Two unit vectors inclined at an angle θ

Proof:

Since vectors are unit vectors

Now,

= 1+1+2×1×1×cosθ

= 2 + 2cosθ

= 2(1 + cosθ)

= 1+1-2×1×1×cosθ

= 2 - 2cosθ

= 2(1 - cosθ)

Dividing above by result (i) we will get,

Proved

The sum of two unit vectors is a unit vector

Calculation:

Also,

Now squaring both sides we get

Now,

Using the above value,

= 3

Hence, the magnitude of their difference is √3.

Given Data:

Three mutually perpendicular unit vectors

Calculation:

= 1+1+1

= 3

Given Data:

Calculation:

…Eq. 1

Now,

= 1600

…Eq. 2

Adding Eq. 1 and Eq. 2

Calculation:

Angle with x-axis

Angle with y-axis

Angle with z-axis

Hence α = β = γ.

Given Data:

Similarly,

Hence these vectors are mutually perpendicular.

Squaring both sides

Given Data:

For this vector to be ⊥

Given Data:

Also,

For this vector to be ⊥

6×4 + (λ+3)(λ-3) -16=0

24+ λ² – 9 - 16=0

λ² = 1

λ = ±1

Given Data:

Now

Also,

3(2-3λ) + 4(1-4λ) - 5(4+5λ) = 0

-50λ = 10

Using the above value,

Similarly,

Given Vectors:

First show that the vectors form a triangle, so we use the addition of vector

Hence these vectors form a triangle

Now we will use Pythagoras theorem to prove this is a right angle triangle.

Therefore these vectors form a right angled triangle.

Given Data:

For this vector to be ⊥

3(-λ+2)+1(2λ+2)=0

-λ + 8 = 0

λ = 8

The value of λ is 8.

Given Data:

Now the angle A

Now the angle B

Now the sum of angles of a triangle is π

∴ A + B + C = π

∴

Given Data:

Magnitude of vectors is unity.

Given Data:

= -2×3+(-1)×(-2)+1×4

= -6+2+4

Angle A right angle, ABC is right angle triangle.

Given Data:

Now the angle B

Given Data:

Now the angle C

So angle C is a right angle triangle.

we know that where x is the angle between two vectors, so gives the projection of vector b on a

Now applying the formula for projection of on

Substituting these values in above formula, we get

meaning of orthogonal is that two vectors are perpendicular to each other, so their dot product is zero.

Similarly,

So, to satisfy the orthogonal condition

Hence proved

Assume,

Using formula:

since it is a unit vector

First taking dot product with

Taking dot product with

Now we have as

Since the magnitude of is 1

Considering,

Therefore, angle with is

Expanding the given equation , we get,

6(2)² + 11(1) – 35(1)²

24 + 11 – 35 = 0

Hence, .

(i) Expanding the given equation

as given

(ii) expanding the given equation

as given

(i) expanding the given equation

Substituting

(ii) expanding the given equation

Substituting,

(iii) expanding the given equation

Substituting,

(i) using formula,

Substituting the given values in above equation we get,

(ii) using formula,

Substituting the given values in above equation we get,

(iii) using formula,

Substituting the given values in above equation we get,

Now this will yield imaginary value.

We know that,

Therefore,

(i) we know that where x is the angle between two vectors

x = 45°

(ii) we know that,

Where, x is the angle between two vectors.

let where u is vector parallel to b and v is vector perpendicular to b, as given in the question.

So, ; where p is some constant

Substituting this value in above equation

Now according to conditions since vector v and b are perpendicular to each other

(5 – 3p) (3) +(5 – p) =0

15 – 9p + 5 – p = 0

20 = 10p

P = 2

So,

substituting this value in above equation, we will get

Let

The angle between these vectors is 30°

So, applying the formula,

So, the magnitude of

Let and

let where u is vector parallel to b and v is vector perpendicular to b.

So, ; where p is some constant

Substituting this value in above equation

Now according to conditions since vector v and b are perpendicular to each other

2(2 – 2p) – 4(1 + 4p) – 2(3 + 2p) = 0

4 – 4p – 4 – 16 p – 6 – 4p = 0

– 24 p = 6

Substituting this value of u vector in above equation

let and

let where u is vector parallel to b and v is vector perpendicular to b

So, ; where p is some constant

Substituting this value in above equation

Now according to conditions since vector v and b are perpendicular to each other

6 – p – 3 – p – 6 – p = 0

P = –1

So,

Substituting this value of in above equation

Meaning of orthogonal is that two vectors are perpendicular to each other, so their dot product is zero.

Similarly

So, to satisfy the orthogonal condition

it is given that and

From this, we can say that

So is a zero vector

And from the second part we can say that can be any vector perpendicular to zero vector .

It is given that is perpendicular to both and

So, and

For to be perpendicular to ,

For the second part.

For to be perpendicular to ,

Hence, proved

we know that

Now expanding LHS of given equation we get,

Taking LCM we get,

Using re-writing the above equation

Hence, proved.

Given that are non-coplanar and and

From above given conditions we can say that either

(i) or

(ii) is perpendicular to and

Since and are non-coplanar, cannot be simultaneously perpendicular to all three, as only three axes exist that is x, y, z

So must be a null vector which is equal to 0

Given is perpendicular to and ,so and

Let a random vector in the plane of where p and k are some arbitrary constant

Taking dot product of

Using and

Hence, proved………

Given

Now squaring both sides, using,

we get,

Hence, proved.

Given

Now squaring both sides using:

We know that,

Where, x is the angle between two vectors

Applying for

Since angle between is acute cos x should be greater than 0

x² – 4 > 0

x > 2 and x < –2

applying for

Since angle between is obtuse cos x should be less than

0

x² – 9 < 0

x > –3 and x < 3

given is perpendicular to so

Applying,

6 + x – y = 0

X – y = –6…(i)

Since the magnitude of both vectors are equal

(y–x) (y+x) =5

6x+6y=5…(ii)

Solving equation (i) and (ii) we get

Given and

Squaring both sides

Now expanding the equation

Given

Squaring both sides we get,

Hence, proved.

Given:- , P and Q are trisection of AB

i.e. AP = PQ = QB or 1:1:1 division of line AB

To Prove:-

Proof:- Let be position vector of O, A and B respectively

Now, Find position vector of P, we use section formulae of internal division: Theorem given below

“Let A and B be two points with position vectors

respectively, and c be a point dividing AB internally in the ration m:n. Then the position vector of c is given by

By above theorem, here P point divides AB in 1:2, so we get

⇒

⇒

Similarly, Position vector of Q is calculated

By above theorem, here Q point divides AB in 2:1, so we get

⇒

⇒

Length OA and OB in vector form

⇒

⇒

⇒

⇒

Now length/distance OP in vector form

⇒

⇒

⇒

⇒

⇒

length/distance OQ in vector form

⇒

⇒

⇒

⇒

⇒

Taking LHS

OP² + OQ²

=

=

as we know in case of dot product

Angle between OA and OB is 90°,

⇒

⇒

Therefore, OP² + OQ²

=

=

=

=

As from figure OA² + OB² = AB²

=

= RHS

Hence, Proved.

Given:- Quadrilateral OACB with diagonals bisect each other at 90°.

Proof:-It is given diagonal of a quadrilateral bisect each other

Therefore, by property of parallelogram (i.e. diagonal bisect each other) this quadrilateral must be a parallelogram.

Now as Quadrilateral OACB is parallelogram, its opposite sides must be equal and parallel.

⇒ OA = BC and AC = OB

Let, O is at origin.

are position vector of A and B

Therefore from figure, by parallelogram law of vector addition

And, by triangular law of vector addition

As given diagonal bisect each other at 90°

Therefore AB and OC make 90° at their bisecting point D

⇒

Or, their dot product is zero

⇒

⇒

⇒

⇒

⇒

Hence we get

OA = AC = CB = OB

i.e. all sides are equal

Therefore by property of rhombus i.e

Diagonal bisect each other at 90°

And all sides are equal

Quadrilateral OACB is a rhombus

Hence, proved.

Given:- Right angle Triangle

To Prove:- Square of the hypotenuse is equal to the sum of the squares of the other two sides

Let ΔAOB be right angle triangle with right angle at O

Thus we have to prove

AB² = OA² + OB²

Proof: - Let, O at Origin, then

Since OB is perpendicular at OA, their dot product equals to zero

)

⇒

⇒ ……(i)

Now,

We can see that, by triangle law of vector addition,



Therefore,

⇒

From equation (i)

⇒

⇒ (Pythagoras theorem)

Hence, proved.

Given:- Parallelogram OABC

To Prove:- AC² + OB² = OA² + AB² + BC² + CO²

Proof:- Let, O at origin

Therefore,

Distance/length of AC

By triangular law:-

the the vectors form sides of triangle

⇒

As AB = OC and BC = OA

From figure

⇒

⇒ ……(i)

Similarly, again from figure

⇒

⇒

⇒

⇒ ……(ii)

Now,

Adding equation (i) and (ii)

⇒ ……(iii)

Take RHS

OA² + AB² + BC² + CO²

……(iv)

Thus from equation (iii) and (iv), we get

LHS = RHS

Hence proved

Given:- ABCD is a rectangle

To prove:- PQRS is rhombus thus finding its properties in PQRS

i.e. All sides equal and parallel

Let, P, Q, R and S are midpoints of sides AB, BC, CD and DA respectively

Therefore

also AB = CD, BC = AD (ABCD is rectangle opposite sides are equal)

Therefore

AP = PB = DR = RC and BQ = QC = AS = SD ……(i)

IMP:- Direction/arrow head of vector should be placed correctly

Now, considering in vector notion and applying triangular law of vector addition, we get

⇒

⇒

⇒

⇒

Magnitude PQ = AC

and

⇒

⇒

⇒

Magnitude SR = AC

Thus sides PQ and SR are equal and parallel

It shows PQRS is a parallelogram

Now,

⇒

⇒

⇒

By Dot product, we know

Here ABCD is rectangle and have 90° at A, B, C, D

⇒

And

⇒

again by triangular law

⇒

⇒

By Dot product, we know

Here ABCD is rectangle and have 90° at A, B, C, D

⇒

From above similarities of sides of rectangle in eq (i), we have

⇒

Hence PQ = PS

And from above results we have

All sides of parallelogram are equal

PQ = QR = RS = SP

Hence proved by property of rhombus (all sides are equal and opposite sides are parallel), PQRS is rhombus

Given:- Rhombus OABC i.e all sides are equal

To Prove:- Diagonals are perpendicular bisector of each other

Proof:- Let, O at the origin

D is the point of intersection of both diagonals

be position vector of A and C respectively

Then,

Now,

⇒

as AB = OC

⇒ ……(i)

Similarly

⇒

⇒ ……(ii)

Tip:- Directions are important as sign of vector get changed

Magnitude are same AC = OB = √a² + c²

Hence from two equations, diagonals are equal

Now let’s find position vector of mid-point of OB and AC

⇒

⇒

and

⇒

⇒

Magnitude is same AD = DC = OD = DB = 0.5(√a² + c²)

Thus the position of mid-point is same, and it is the bisecting point D

By Dot Product of OB and AC vectors we get,

⇒

⇒

⇒

⇒

As the side of a rhombus are equal OA = OC

⇒

⇒

Hence OB is perpendicular on AC

Thus diagonals of rhombus bisect each other at 90°

Given:- ABCD is a rectangle i.e AB = CD and AD = BC

To Prove:- ABCD is a square only if its diagonal are perpendicular

Proof:- Let A be at the origin

Now,

By parallelogram law of vector addition,

⇒

Since in rectangle opposite sides are equal BC = AD

⇒

⇒

and

⇒

Negative sign as vector is opposite

⇒

⇒

Diagonals are perpendicular to each other only

⇒

⇒

⇒

⇒

⇒ AB² = AD²

⇒ AB = AD

Hence all sides are equal if diagonals are perpendicular to each

other

ABCD is a square

Hence proved

Given:- ΔABC and AD is median

To Prove:- AB² + AC² = 2(AD² + CD²)

Proof:- Let, A at origin

be position vector of B and C respectively

Therefore,

Now position vector of D, mid-point of BC i.e divides BC in 1:1.

Section formula of internal division: Theorem given below

“Let A and B be two points with position vectors

respectively, and c be a point dividing AB internally in the ration m:n. Then the position vector of c is given by

Position vector of D is given by

⇒

Now distance/length of CD

= position vector of D-position vector of C

⇒

⇒

Now taking RHS

= 2(AD² + CD²)

=

=

=

=

=

= AB² + AC²

= LHS

Hence proved

Given:- ΔABC, AD is median

To Prove:- If AD is perpendicular on base BC then ΔABC is isosceles

Proof:- Let, A at Origin

be position vector of B and C respectively

Therefore,

Now position vector of D, mid-point of BC i.e divides BC in 1:1

Section formula of internal division: Theorem given below

“Let A and B be two points with position vectors

respectively, and c be a point dividing AB internally in the ration m:n. Then the position vector of c is given by

Position vector of D is given by

⇒

Now distance/length of BC

= position vector of C-position vector of B

⇒

Now, assume median AD is perpendicular at BC

Then by Dot Product

⇒

⇒

⇒

⇒

⇒

⇒ AC = AB

Thus two sides of ΔABC are equal

Hence ΔABC is isosceles triangle

Given:- Quadrilateral ABCD with AC and BD are diagonals. P and Q are mid-point of AC and BD respectively

To Prove:- AB² + BC² + CD² + DA² = AC² + BD² + 4PQ²

Proof:- Let, O at Origin

be position vector of A, B, C and D respectively

As P and Q are mid-point of AC and BD,

Then, position vector of P, mid-point of AC i.e divides AC in 1:1

and position vector of Q, mid-point of BD i.e divides BD in 1:1

Section formula of internal division: Theorem given below

“Let A and B be two points with position vectors

respectively, and c be a point dividing AB internally in the ration m:n. Then the position vector of c is given by “

Hence

Position vector of P is given by

Position vector of Q is given by

Distance/length of PQ

⇒

⇒

Distance/length of AC

⇒

⇒

Distance/length of BD

⇒

⇒

Distance/length of AB

⇒

⇒

Distance/length of BC

⇒

⇒

Distance/length of CD

⇒

⇒

Distance/length of DA

⇒

⇒

Now, by LHS

= AB² + BC² + CD² + DA²

Where are angle between vectors

Take RHS

AC² + BD² + 4PQ²

Thus LHS = RHS

Hence proved

We know,

where θ is the angle between a⃗ and b⃗.

Given, |a|=2 |b|=√3

So,

θ = 60°

Given,

|a|=3, |b|=4

We know,

If A⃗ = , B⃗ =

And

So,

|A|= √34

|B|= √6

Here, 4×2 + (-3) × (-1) + 3×(-1) = 8

Orthogonal vectors are perpendicular to each other so their dot product is always 0 as cos90°=0

If A⃗ = , B⃗ =

And

6-m-8=0

-m-2=0

m=-2

If A⃗ = , B⃗ =

And A⃗ is parallel to B⃗

Then A⃗= kB⃗, where k is some constant

So,

m=6×4

=24

We know that dot product is distributive.

So

We know

=0

We know that dot product is distributive.

So

Given that,

Therefore, both the vectors have equal magnitude

We know,

Comparing LHS and RHS we can conclude that

We know,

If

Then,

Comparing LHS and RHS we can conclude that

Given,

and

(as magnitude cannot be negative)

Possible answers are,

i.e. is a null vector

Or

or i.e. and are perpendicular

Or

i.e. is a null vector

(As given as unit vector)

Projection of on is

Therefore projection =

= 2

x-axis=

y-axis=

z-axis=

Projection along x-axis

Projection along y-axis

Projection along z-axis

Component of a given vector along is given by the length of on .

Let be the angle between both the vectors.

So the length of on is given as:

By vector dot product, we know that:

Cos =

Therefore,

Hence,

Let

(1)

(2)

(3)

Put the values obtained in the given equation

We get:

=

i.e.

=

Given:

(perpendicular)

So,

=0

Therefore;

sin

Multiply and divide the whole equation by 2:

We get

By the identity:

We have:

So

, n

Let, &

We know that, projection of along is given by:

Also,

&

So,

Let +

For all the equations to be equal to 1;

i.e. x= x + y

= x + y + z

=1

So, x=1;

& x + y=1

& x + y + z=1

We get: x=1,y=z=0

Therefore,

Since,

Let &

Angle between is and angle between is

By vector addition method;

we have:

=2(1+cos θ)

=2(1-cos θ)

So,

Now in the parallelogram:

Area of parallelogram= (product of two sides and the sine of angle between them)

i.e. (1)

Also area of parallelogram= sum of area of all four triangle

And area of each triangle =

So, Area = 2

Since are supplementary

A= 4= (2)

From (1) &(2) we get:

Since are mutually perpendicular;

Then, cos (1)

And sin (2)

Squaring and adding both equations, we get;

1=

So,

Hence,

Since all three vectors are mutually perpendicular, so dot product of each vector with another is zero.

i.e.

Also,

So,

i.e.

So,

By vector dot product, we know that:

So, cos=

= &

Therefore,

cos =

cos=

So,

Let

for to be perpendicular to

i.e. [vector dot product]

(2).()=0

2-2+3=0

5-2=0

Hence, =

We know that;

So,

for to be parallel to

i.e. ( [vector cross product]

= 0

=0

=0

(6-9p)=0 & (3p-2)=0

Hence,

Let and

for to be perpendicular to

i.e. [vector dot product]

6+2λ-12=0

2λ-6=0

Hence, λ=3

Given and and

The projection of vector on a is given by,

(since scalar product is commutative)

We know that the scalar product of two non-zero vectors and , denoted by , is defined as,

=

Let and

Then the projection of vector on is given by,

=

=1× 2-3× 3+7× 6

=2-9+42

=35

Now, |

=

=7

Therefore projection of on

=5

Given and

Projection of on is given by

=2+6+12

=2λ+18

Now, |b|= ==7

2λ+18=28

2λ=10

λ=5

Given and

For two vectors to be perpendicular, the angle between them must be or

We know that cos 90=0

=2-2λ+3

=5-2λ

By scalar product,

5-2λ=0

Let and

Projection of on is given by,

=7×2+1× 6 – 4× 3

=14+6-12

=14-6

=8

Therefore, projection of is

Given and

For two vectors to be perpendicular, the angle between them must be or

We know that Cos90=0

=2-2λ+3

=5-2λ

By scalar product,

Therefore, 5-2λ=0

Given,

So,

Now, to find projection of

Now,

=3× 2-1× 2+6× 1

=6-2+6

=10

=√9

=3

Therefore,

Given, | and |

Also given and are perpendicular

Given |

Also given, is a unit vector

By vector product,

Therefore,

Given |and

Now,

Also,

Therefore,

We know that and cos is negative in 2^nd quadrant

Therefore, θ =180-60

=120

Given, and

By scalar product,

By substituting the values, we get

=1

3a² -2√3 cos θ + b² =1

⇒3-2√3 cos θ +1=1

⇒4-1=2√3 cos θ

⇒3=2√3 cos θ

Here, and

Also, and

and

Solving above two equations for and we get,

and

Also,

Ands,

Now, θ is the angle between and

So,

Here, __________(1)

_________(2)

and _________(3)

From (2),

(____________(4)

From (3) and (4)

()

)

So,

Here, _____________(1)

() ___________(2)

From (1)

() _________(3)

From (1)

() __________(4)

From (2) and (3)

____________(5)

From (2) and (5)

Let θ be the angle between and

Then,

So,

i.e. angle between and is .

Here, and are unit vectors.

i.e. and

If is unit vector then

()

( ;; )

Now,

We know and cosine is negative in second quadrant.

Let

So,

=cos² α(cos² β+sin² β)+sin² α

=cos ²α(1)+sin²α

=1

i.e.

So, is a unit vector.

Let be the direction of

Then,

Let θ be the angle between and Y-axis

Then ,

Here, and are unit vectors.

i.e. and

Now, Let θ be the angle between and

So,

Now, we know

Therefore, is not possible.

Let and

Given that and are perpendicular.

So,

⇒1-4x² -6y² =0

⇒4x²+6y²=1

Here, vectors are in 3-Dimensions

∴ above equation represents an ellipse .i.e. locus of (x, y) is an ellipse.

Let be the vector projection of onto

Then,

Now, vector component of perpendicular to is

Here, and

The parallelogram is constructed on and

Then its one diagonal is

And other diagonal is

Length of one diagonal is

=√225

=15

Length of other diagonal is

=√593

So, Length of the longest diagonal is √593.

Here,

Now,is unit vector if

i.e.

i.e. |

i.e.

Here, θ be the angle between and

Then,

Now,

⇒cosθ≥0

We know cosine is positive in first quadrant.

Here, angle between and is obtuse

So,

⇒ 14x² – 8x + x ≤ 0

⇒ 14x² – 7x ≤ 0

⇒ 2x² – x ≤ 0

⇒ x(2x – 1)≤ 0

or __________(1)

Now, angle between and Z-axis is acute

So,

⇒x≥0 ___________(2)

∴ From (1) and (2) .

We know that,

(i)

Since, they are mutually perpendicular vectors

===0 (ii)

And according to question

Using (i) and (ii) ,

Ans.

Smart Approach

In case of such mutually perpendicular vectors, assume vectors to be and verify your answer from options.

We have,

=++

=-+8

Given that and are perpendicular

.=0

.=0

6-+8=0

Ans.

Projection of on is

Projection of on is

= 1 Ans.

The given two vectors,

Their dot-product is zero

2a+3b-4c=0

From the given options only option B satisfies the above equation

Hence option B is correct answer.

=0Ans.

Given that,

Ans.

If are unit vector then

Maximum value of a sin θ +bcosΘ is

Maximum value of + 2 is 4 Ans.

=

If the angle is acute

<0

(+3).()<0

⟹ x²-3x-28<0

(x-7) (x+4)<0

xR-(4,7) Ans.

We know that,

If and are two-unit vectors inclined at an angle θ

According to question,

<1

Ans.

We know that,

(i)

Since,

is perpendicular to

=0

And according to question

=1

We can rewrite equation (i) as

1=1+1+1+0+2(cos α + cosβ)

cos α + cosβ = -1 Ans.

Key Concept/Trick: Magnitude of Projection of any vector

is given by

Now, Since it is the magnitude or length(a) we have to give the length a direction in the direction of

So, we multiply the projection by unit vector of

). which on simplification gives option B Ans.

Since, the given two vectors are given as perpendicular their dot product must be zero

= 0

tan=

Since is acute then, = Ans

We know that,

=

According to Question,

=1

=

= Ans.