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NCERT solution class 12 chapter 4 Differential Equations exercise 4.5 mathematics part 2

EXERCISE 4.5


Page No 458:

Question 1:

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

Answer:

If is a unit vector in the XY-plane, then 

Here, θ is the angle made by the unit vector with the positive direction of the x-axis.

Therefore, for θ = 30°:

Hence, the required unit vector is.

Question 2:

Find the scalar components and magnitude of the vector joining the points

.

Answer:

The vector joining the pointscan be obtained by,

Hence, the scalar components and the magnitude of the vector joining the given points are respectively and.

Question 3:

A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

Let O and B be the initial and final positions of the girl respectively.

Then, the girl’s position can be shown as:

Now, we have:

By the triangle law of vector addition, we have:

Hence, the girl’s displacement from her initial point of departure is

.

Question 4:

If, then is it true that? Justify your answer.

Answer:

Now, by the triangle law of vector addition, we have.

It is clearly known that  represent the sides of ΔABC.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

Hence, it is not true that.

Question 5:

Find the value of x for whichis a unit vector.

Answer:

is a unit vector if.

Hence, the required value of x is.

Question 6:

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

.

Answer:

We have,

Letbe the resultant of.

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors is

Question 7:

If, find a unit vector parallel to the vector.

Answer:

We have,

Hence, the unit vector alongis

Question 8:

Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).

Thus, the given points A, B, and C are collinear.

Now, let point B divide AC in the ratio. Then, we have:

On equating the corresponding components, we get:

Hence, point B divides AC in the ratio

Question 9:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors areexternally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

It is given that.

It is given that point R divides a line segment joining two points P and Q externally in the ratio 1: 2. Then, on using the section formula, we get:

Therefore, the position vector of point R is.

Position vector of the mid-point of RQ =

Hence, P is the mid-point of the line segment RQ.

Question 10:

The two adjacent sides of a parallelogram areand .

Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

Adjacent sides of a parallelogram are given as: and

Then, the diagonal of a parallelogram is given by.

Thus, the unit vector parallel to the diagonal is

Area of parallelogram ABCD =

Hence, the area of the parallelogram issquare units.

Question 11:

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are.

Answer:

Let a vector be equally inclined to axes OX, OY, and OZ at angle α.

Then, the direction cosines of the vector are cos α, cos α, and cos α.

Hence, the direction cosines of the vector which are equally inclined to the axes are.

Question 12:

Let and. Find a vector which is perpendicular to both and, and.

Answer:

Let.

Sinceis perpendicular to bothand, we have:

Also, it is given that:

On solving (i), (ii), and (iii), we get:

Hence, the required vector is.

Question 13:

The scalar product of the vectorwith a unit vector along the sum of vectors and is equal to one. Find the value of.

Answer:

Therefore, unit vector alongis given as:

Scalar product ofwith this unit vector is 1.

Hence, the value of λ is 1.

Question 14:

If are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to and.

Answer:

Sinceare mutually perpendicular vectors, we have

It is given that:

Let vector be inclined to at angles respectively.

Then, we have:

Now, as.

Hence, the vectoris equally inclined to.

Page No 459:

Question 15:

Prove that, if and only if are perpendicular, given.

Answer:

Question 16:

If θ is the angle between two vectors and , then only when

(A)  (B) 

(C)  (D) 

Answer:

Let θ be the angle between two vectors and.

Then, without loss of generality, and are non-zero vectors so that.

It is known that.

Hence,  when.

The correct answer is B.

Question 17:

Let  and  be two unit vectors andθ is the angle between them. Then is a unit vector if

(A)  (B)

 (C)

 (D)

Answer:

Let  and  be two unit vectors andθ be the angle between them.

Then, .

Now,  is a unit vector if.

Hence, is a unit vector if.

The correct answer is D.

Question 18:

The value of is

(A) 0 (B) –1 (C) 1 (D) 3

Answer:

The correct answer is C.

Question 19:

If θ is the angle between any two vectors and, then when θisequal to

(A) 0 (B)  (C)  (D) π

Answer:

Let θ be the angle between two vectors and.

Then, without loss of generality, and are non-zero vectors, so that.

Hence, when θisequal to.

The correct answer is B.


 

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