Vector Algebra

EXERCISE 10.1

Question 1:

Represent graphically a displacement of 40 km, 30° east of north.

ANSWER:

Here, vector represents the displacement of 40 km, 30° East of North.

Question 2:

Classify the​​ following measures as scalars and vectors.

(i) 10 kg (ii) 2 metres north-west (iii) 40°

(iv) 40 watt (v) 10–19 coulomb (vi) 20 m/s2

ANSWER:

(i) 10 kg is a scalar quantity because it involves only magnitude.

(ii) 2 meters north-west is a vector​​ quantity as it involves both magnitude and direction.

(iii) 40° is a scalar quantity as it involves only magnitude.

(iv) 40 watts is a scalar quantity as it involves only magnitude.

(v) 10–19 coulomb is a scalar quantity as it involves only magnitude.

(vi)​​ 20 m/s2 is a vector quantity as it involves magnitude as well as direction.

Question 3:

Classify the following as scalar and vector quantities.

(i) time period (ii) distance (iii) force

(iv) velocity (v) work done

ANSWER:

(i) Time period is a​​ scalar quantity as it involves only magnitude.

(ii) Distance is a scalar quantity as it involves only magnitude.

(iii) Force is a vector quantity as it involves both magnitude and direction.

(iv) Velocity is a vector quantity as it involves both magnitude​​ as well as direction.

(v) Work done is a scalar quantity as it involves only magnitude.

Question 4:

In Figure, identify the following vectors.

(i) Coinitial (ii) Equal (iii) Collinear but not equal

ANSWER:

(i) Vectors and are coinitial because​​ they have the same initial point.

(ii) Vectorsandare equal because they have the same magnitude and direction.

(iii) Vectorsand are collinear but not equal. This is because although they are parallel, their directions are not the same.

Question 5:

Answer the following as true or false.

(i)  andare collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

(iv) Two collinear vectors having the same magnitude are equal.

ANSWER:

(i) True.

Vectors  andare parallel to the same line.

(ii) False.

Collinear vectors are those vectors that are parallel to the same line.

(iii) False.

It is not necessary for two vectors having the same magnitude to be parallel to the same line.

(iv) False.

Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.

EXERCISE 10.2

Question 1:

Compute the magnitude of the following vectors:

ANSWER:

The given vectors​​ are:

Question 2:

Write two different vectors having same magnitude.

ANSWER:

Hence, are two different vectors having the same magnitude. The vectors are different because they have different directions.

Question 3:

Write two different vectors​​ having same direction.

ANSWER:

The direction cosines of are the same. Hence, the two vectors have the same direction.

Question 4:

Find the values of x and y so that the vectors are equal

ANSWER:

The two vectors will be equal if their​​ corresponding components are equal.

Hence, the required values of x and y are 2 and 3 respectively.

Question 5:

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

ANSWER:

The vector with the​​ initial point P (2, 1) and terminal point Q (–5, 7) can be given by,

Hence, the required scalar components are –7 and 6 while the vector components are 

Question 6:

Find the sum of the vectors.

ANSWER:

The given vectors are.

Question 7:

Find​​ the unit vector in the direction of the vector.

ANSWER:

The unit vector in the direction of vector is given by.

Question 8:

Find the unit vector in the direction of vector, where P and Q are the points

(1, 2, 3) and (4, 5, 6),​​ respectively.

ANSWER:

The given points are P (1, 2, 3) and Q (4, 5, 6).

Hence, the unit vector in the direction of  is

.

Question 9:

For given vectors, and , find the unit vector in the direction of the vector 

ANSWER:

The given vectors are and.

Hence, the unit vector in the direction of is
(a→+b→)∣∣∣a→+b→∣∣∣=iˆ+kˆ2√=12√i⌢+12√k⌢.a→+b→a→+b→=i^+k^2=12i⏜+12k⏜.

Question 10:

Find a vector in the direction of vector which has magnitude 8 units.

ANSWER:

Hence, the vector in the direction of​​ vector which has magnitude 8 units is given by,

Question 11:

Show that the vectorsare collinear.

ANSWER:

.

Hence, the given vectors are collinear.

Question 12:

Find the direction cosines of the vector 

ANSWER:

Hence, the direction​​ cosines of 

Question 13:

Find the direction cosines of the vector joining the points A (1, 2, –3) and

B (–1, –2, 1) directed from A to B.

ANSWER:

The given points are A (1, 2, –3) and B (–1, –2, 1).

Hence, the direction cosines of are 

Question 14:

Show that the vector is equally inclined to the axes OX, OY, and OZ.

ANSWER:

Therefore, the direction cosines of 

Now, let α, β, and γbe the angles formed by with the positive directions of x, y, and z axes.

Then, we have

Hence, the​​ given vector is equally inclined to axes OX, OY, and OZ.

Question 15:

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  respectively, in the ration 2:1

(i) internally

(ii)​​ externally

ANSWER:

The position vector of point R dividing the line segment joining two points

P and Q in the ratio m: n is given by:

• Internally:

• Externally:

Position vectors of P and Q are given as:

(i) The position vector of point R which divides the line joining two points P and Q internally in the ratio 2:1 is given by,

(ii) The position vector of point R which divides the line joining two points P and Q externally in the ratio 2:1 is given by,

Question 16:

Find the position vector of the​​ mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).

ANSWER:

The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, – 2) is given by,

Question 17:

Show that the points A, B and C with​​ position vectors,, respectively form the vertices of a right angled triangle.

ANSWER:

Position vectors of points A, B, and C are respectively given as:



∣∣∣AB−→−∣∣∣2+∣∣∣CA−→−∣∣∣2=35+6=41=∣∣∣BC−→∣∣∣2AB→2+CA→2=35+6=41=BC→2

Hence, ABC is a right-angled​​ triangle.

Question 18:

In triangle ABC which of the following is not true:

A. 

B. 

C. 

D. 

ANSWER:

On applying the triangle law of addition in the given triangle, we have:

From equations (1) and (3), we have:

Hence, the equation given in​​ alternative C is incorrect.

The correct answer is C.

Question 19:

If are two collinear vectors, then which of the following are incorrect:

A. , for some scalar λ

B. 

C. the respective components of are proportional

D. both the vectors have same​​ direction, but different magnitudes

ANSWER:

If are two collinear vectors, then they are parallel.

Therefore, we have:

 (For some scalar λ)

If λ = ±1, then .

Thus, the respective components of are proportional.

However, vectors can have​​ different directions.

Hence, the statement given in D is incorrect.

The correct answer is D.

EXERCISE 10.3

Question 1:

Find the angle between two vectorsandwith magnitudesand 2, respectively having.

ANSWER:

It is given that,

Now, we know that.

Hence, the angle between the given vectors andis.

Question 2:

Find the angle between the vectors

ANSWER:

The given vectors are.

Also, we know that.

Question 3:

Find the projection of the vectoron the vector.

ANSWER:

Letand.

Now,​​ projection of vectoronis given by,

Hence, the projection of vector onis 0.

Question 4:

Find the projection of the vectoron the vector.

ANSWER:

Letand.

Now, projection of vectoronis given by,

Question 5:

Show that each of the given​​ three vectors is a unit vector:

Also, show that they are mutually perpendicular to each other.

ANSWER:

Thus, each of the given three vectors is a unit vector.

Hence, the given three vectors are mutually perpendicular to each other.

Question 6:

Findand, if.

ANSWER:

Question 7:

Evaluate the product.

ANSWER:

Question 8:

Find the magnitude of two vectors, having the same magnitude and such that the angle between them is 60° and their scalar product is.

ANSWER:

Let θ be​​ the angle between the vectors

It is given that

We know that.

Question 9:

Find, if for a unit vector.

ANSWER:

Question 10:

Ifare such thatis perpendicular to, then find the value of λ.

ANSWER:

Hence, the required value of λ is 8.

Question 11:

Show that is perpendicular to, for any two nonzero vectors

ANSWER:

Hence, andare perpendicular to each other.

Question 12:

If, then what can be concluded about the vector?

ANSWER:

It is given that.

Hence, vectorsatisfyingcan be any vector.

Question 13:

If are unit vectors such that , find the value of .

ANSWER:

It is given that .

From (1), (2) and (3),

Question 14:

If either vector, then. But the converse need not be true. Justify your​​ answer with an example.

ANSWER:

We now observe that:

Hence, the converse of the given statement need not be true.

Question 15:

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find​​ ∠ABC. [∠ABC is​​ the angle between the vectorsand]

ANSWER:

The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2).

Also, it is given that​​ ∠ABC is the angle between the vectorsand.

Now, it is known that:

.

Question 16:

Show that the​​ points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

ANSWER:

The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).

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

Question 17:

Show that the vectorsform the vertices of a​​ right angled triangle.

ANSWER:

Let vectors be position vectors of points A, B, and C respectively.

Now, vectorsrepresent the sides of ΔABC.

Hence, ΔABC is a right-angled triangle.

Question 18:

Ifis a nonzero vector of magnitude ‘a’ and λ a​​ nonzero scalar, then λis unit vector if

(A) λ = 1 (B) λ = –1 (C) 

(D) 

ANSWER:

Vectoris a unit vector if.

Hence, vectoris a unit vector if.

The correct answer is D.

EXERCISE 10.4

Question 1:

Find, if and.

ANSWER:

We have,

and

Question 2:

Find a unit vector perpendicular to each of the vector and, where and.

ANSWER:

We have,

and

Hence, the unit vector perpendicular to each of the vectors and is given by the relation,