NCERT Solutions for Class 12 Science Math Chapter 4 - Vector Algebra

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Answer:

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

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.

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.

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.

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.

Answer:

The given vectors are:

Answer:

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

Answer:

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

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.

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

Answer:

The given vectors are.

Answer:

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

Answer:

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

Hence, the unit vector in the direction of is

.

Answer:

The given vectors are and.

Hence, the unit vector in the direction of is
$\frac{\left(\overrightarrow{a}+\overrightarrow{b}\right)}{\left|\overrightarrow{a}+\overrightarrow{b}\right|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\stackrel{⏜}{i}+\frac{1}{\sqrt{2}}\stackrel{⏜}{k}.$

Answer:

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

Answer:

.

Hence, the given vectors are collinear.

Answer:

Hence, the direction cosines of

Answer:

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

Hence, the direction cosines of are

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.

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:

1. Internally:

2. 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,

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,

Answer:

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



${\left|\overrightarrow{\mathrm{AB}}\right|}^2+{\left|\overrightarrow{\mathrm{CA}}\right|}^2=35+6=41={\left|\overrightarrow{\mathrm{BC}}\right|}^2$

Hence, ABC is a right-angled triangle.

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.

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.

Answer:

It is given that,

Now, we know that.

Hence, the angle between the given vectors andis.

Answer:

The given vectors are.

Also, we know that.

Answer:

Letand.

Now, projection of vectoronis given by,

Hence, the projection of vector onis 0.

Answer:

Letand.

Now, projection of vectoronis given by,

Answer:

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

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

Answer:

Answer:

Answer:

Let θ be the angle between the vectors

It is given that

We know that.

Answer:

Answer:

Hence, the required value of λ is 8.

Answer:

Hence, andare perpendicular to each other.

Answer:

It is given that.

Hence, vectorsatisfyingcan be any vector.

Answer:

It is given that .

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

Answer:

We now observe that:

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

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:

.

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.

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.

Answer:

Vectoris a unit vector if.

Hence, vectoris a unit vector if.

The correct answer is D.

Answer:

We have,

and

Answer:

We have,

and

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

Answer:

Let unit vector have (a₁, a₂, a₃) components.

⇒

Since is a unit vector, .

Also, it is given that makes angleswith with , and an acute angle θ with

Then, we have:

Hence, and the components of are.

Answer:

Answer:

On comparing the corresponding components, we have:

Hence,

Answer:

Then,

(i) Either or, or

(ii) Either or, or

But, and cannot be perpendicular and parallel simultaneously.

Hence, or.

Answer:

We have,

On adding (2) and (3), we get:

Now, from (1) and (4), we have:

Hence, the given result is proved.

Answer:

Take any parallel non-zero vectors so that.

It can now be observed that:

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

Answer:

The vertices of triangle ABC are given as A (1, 1, 2), B (2, 3, 5), and

C (1, 5, 5).

The adjacent sidesand of ΔABC are given as:

Area of ΔABC

Hence, the area of ΔABC

Answer:

The area of the parallelogram whose adjacent sides are is.

Adjacent sides are given as:

Hence, the area of the given parallelogram is.

Answer:

It is given that.

We know that, where is a unit vector perpendicular to both and and θ is the angle between and.

Now, is a unit vector if.

Hence, is a unit vector if the angle between and is.

The correct answer is B.

Answer:

The position vectors of vertices A, B, C, and D of rectangle ABCD are given as:

The adjacent sides and of the given rectangle are given as:

$\Rightarrow \left|\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{BC}}\right|=2$

Now, it is known that the area of a parallelogram whose adjacent sides are is.

Hence, the area of the given rectangle is

The correct answer is C.

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.

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.

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

.

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.

Answer:

is a unit vector if.

Hence, the required value of x is.

Answer:

We have,

Letbe the resultant of.

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

Answer:

We have,

Hence, the unit vector alongis

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

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.

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.

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.

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.

Answer:

Therefore, unit vector alongis given as:

Scalar product ofwith this unit vector is 1.

Hence, the value of λ is 1.

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.

Answer:

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.

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.

Answer:

The correct answer is C.

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.