Class 12 NCERT Solutions- Mathematics Part ii – Chapter 10 – Vector Algebra Miscellaneous Exercise on Chapter 10

In the article, we will solve Miscellaneous Exercise from Chapter 10, “Vector Algebra” in the NCERT. This exercise covers the basics of vectors like scalar and vector components of vectors, section formulas, Multiplication of a Vector by a scalar, etc.

Question 1: Write down a unit vector in XY-plan, making an angle of 30 degree with the positive direction of x-axis

Answer

Let us take as a unit vector in the XY-plan, then

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

Therefore, for =

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 points can be obtained by,

= Position vector of Q-Position vector of P

Hence, the scalar components and magnitude of the vector joining the points are:

Question 3: A girl walks 4 km towards west, then she walks 3 km in a direction 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.

Now, we have

Also, 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:

By the triangle law of vector addition, we have

Also, we know represent the sides of .

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 which is a unit vector.

Answer:

We know is a unit vector if

Now,

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,

Let be the resultant of

Now,

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

Question 7:

Answer:

We have,

Therefore, the unit vector along

=

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).

Hence, 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 2:3

Question 9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are ()and() externally in the ratio 1: 2. Also, show that P is the mid-point of the line segment RQ.

Answer:

Given

Also, point R divides a line segment joining two points P and Q externally in. the ratio 1: 2. By the section formula, we get:

Hence, the positive vector of point R is .

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

Answer:

Two adjacent sides of a parallelogram are:

The diagonal of parallelogram is given by

Thus, the unit vector parallel to the diagonal is

Area of parallelogram ABCD =

Hence, the area of the parallelogram is square units.

Question 11: Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

Answer:

Consider a vector is equally inclined to axes OX, OY, and OZ at angle a.

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

Now

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

Question 12: Find a vector which is perpendicular to both

Answer:

We know is perpendicular to both

Also,

And

Solving (i),(ii),and (iii),we get:

Therefore, the required vector is

Question 13: The scalar product of the vector with a unit vector along the sum of vectors is equal to one. Find the value of λ.

Answer:

Therefore, unit vector along

Scalar product of with its unit vector is 1.

Therefore, the value of is 1.

Question 14: if mutually perpendicular vectors of equal magnitudes are, show that the vector equally inclined to .

Answer:

Given that are mutually perpendicular vectors, Hence we have

Also,

The vector be inclined to at angle

Now we have:

Now, as

Hence, the vector is equally inclined to

Question 15: Prove that ,if only if are perpendicular, given

Answer:

Question 16: if is the angle between two vectors and ,then

Answer:

Let θ be the angle between two vectors

Now =

Correct answer is

Question 17: Let and be two-unit vectors a and is the angle between them.Then is a unit vector if

Answer:

We have two unit vectors with angle θ between them.

Then,

Also if

Hence, Correct answer is

Question 18: The value of

Answer:

Therefore, correct answer is

Question 19: If is the angle between any two vectors , then when is equal to

Answer:

Let θ be the angle between

are non-zero vectors, so

Hence, Correct answer is

Also, Check

• Class 12 NCERT Solutions- Mathematics Part ii – Chapter 10 – Vector Algebra Exercise 10.1
• Class 12 NCERT Solutions – Mathematics Part ii – Chapter 10 – Vector Algebra Exercise 10.2