Courses

# Vector Calculus - 4

## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Vector Calculus - 4

Description
This mock test of Vector Calculus - 4 for Mathematics helps you for every Mathematics entrance exam. This contains 20 Multiple Choice Questions for Mathematics Vector Calculus - 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Vector Calculus - 4 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Vector Calculus - 4 exercise for a better result in the exam. You can find other Vector Calculus - 4 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
QUESTION: 1

### The direction derivative of the scalar function f(x, y, z) = x2 + 2y2 + z at the point P = (1, 1, 2) in the direction of the vector

Solution:

Since,

At (1, 1, 2)

So, direction derivative of at P(1, 1, 2) in direction of

QUESTION: 2

### The maximum value of the directional derivative of the function φ = 2x2 + 3y2 + 5z2 at a point (1, 1, -1) is

Solution:

Hence, vector field is irrotational and divergence-free.
Given,

Directional derivative of φ is
(given)
As x = 1, y = 1, z = - 1

Magnitude of directional derivative is

QUESTION: 3

### A velocity vector is given as  Then, the divergence of this velocity vector at (1, 1, 1) is

Solution:

= 5y + 4y + 6yz

QUESTION: 4

If a vector  has a constant magnitude, then

Solution:

is of constant magnitude.

QUESTION: 5

where P is a vector, is equal to

Solution:

We have from the property of vector triple product

QUESTION: 6

then the value of curl

Solution:

QUESTION: 7

Find the area of a triangle formed by the tips of vectors

Solution:

In figure from A, draw AD ⊥ BC. So, in right angled

So, from area of ΔABC

QUESTION: 8

The angle between two unit magnitude coplanar vectors P(0.866,0.500,0) and Q(0.259,0.966,0) will be

Solution:

We have,

QUESTION: 9

The directional derivative of  in the direction of

Solution:

Given,

Problem is to find directional derivative of
f at A(1, 1) along the vector
We take,

QUESTION: 10

The inner (dot) product o f two non-zero vectors  and is  zero. The angle (degrees) between the two vectors is

Solution:

QUESTION: 11

A sphere of unit radius is centred at the origin. The unit normal at point (x, y, z) on the surface of the sphere is the vector

Solution:

Equation of sphere of unit radius having centre at origin is
f (x, y, z) = x2 + y2 + z2 - 1 = 0
So,

being unit normal at (x,y, z) on the surface.

QUESTION: 12

Let x and y be two vectors in a three-dimensional space and < x, y > denote their dot product. Then the determinant,

Solution:

and let

So,

Now, on putting
D = 0 or

Vector   are linearly dependent.
So, linearly dependent ⇒ D = 0 and for linearly independent ⇒ D ≠ 0
or Positive and negative
We can also see that D = (x2y1 - x1y2)2 cannot be negative.
So, linearly independent ⇒ D is positive.

QUESTION: 13

For an incompressible flow, the x and y components of the velocity vector are  = 3(y | z), where x, y and z are in metre, all velocities are in m/s. Then, the z component of the velocity vector  of the flow for the boundary condition  at z = 0, is

Solution:

Given

(where, vz = 0 at z = 0 and fluid is incompressible) For incompressible fluid flow, equation of continuity is

or
or
So,

QUESTION: 14

The gradient of the scalar fleld f(x, y) = y2 - 4xy at (1,2) is

Solution:

QUESTION: 15

The value of the surface integral  evaluated over the surface of a cube having sides of length a is is unit normal vector)

Solution:

...(i)
S, represents the surface of a cube having sides each of length   being unit normal vector to the surface.
Problem is to evaluate I. We know from Gauss’s divergence theorem

So, using Eq. (ii) for Eq. (i), we get

QUESTION: 16

Value of the integral  where C is the square cut from the first quadrant by the lines x = 1 and y = 1 will be (use Green’s theorem to change the line integral into double integral)

Solution:

From Green’s theorem

So,

QUESTION: 17

Determine the following integral

where,  is the position vector field  and S is the surface of a sphere of radius R.

Solution:

S being surface of a sphere of radius R.

(using Gauss’s divergence theorem)

QUESTION: 18

are three points having coordinates (3, -2, -1), (1, 3,4) and (2, -1, -2) in XY and Z-plane, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by

Solution:

and  Now, we have to find the distance of point P to the plane OQR according to figure.

We draw a perpendicular from point P to plane OQR and let it meet with perpendicular at S(x, y, z).
Then,

Now  is normal to given plane OQR  and it is also normal to
So, vector normal to

It also have the same direction ratios as a vector normal to
Since  is itself normal to

From Eqs. (iv) and (v), the values of x and z putting in Eq. (Hi) respectively, we get y = 0
Then from Eq. (iv), x = 1
and from Eq. (v), z = -2
So, points of 5 is (1, 0, -2).
So,

QUESTION: 19

The area of the loop of Descartes’s Folium x3 + y3 = 3 axy is

Solution:

Putting y = tx, we get the parametric equation of the contour of the folium as
The loop is described as t varies from 0 to
where θ varies

QUESTION: 20

The divergence of a vector field A is always equal to zero, if the vector field A can be expressed as

Solution:

Let

Now, we consider
Div curl B

Thus, curl B = A and div A = 0