NCERT Exemplar: Vectors | Mathematics (Maths) Class 12 - JEE PDF Download

SHORT ANSWER TYPE QUESTIONS

Q.1. Find the unit vector in the direction of sum of vectors

Ans.
Given that

∴ Unit vector in the direction of


Hence, the required unit vector is

Q.2. Iffind the unit vector in the direction of
(i)
(ii)
Ans.
Given that
(i) 
∴ Unit vector in the direction of


Hence, the required unit vector is
(ii)-

∴ Unit vector in the direction of

Hence, the required unit vector is

Q.3. Find a unit vector in the direction ofwhere P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.
Ans.
Given coordinates are P(5, 0, 8) and Q(3, 3, 2)
∴
∴ Unit vector in the direction of


Hence, the required unit vector is

Q.4. If are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.
Ans.
Given that
BC = 1.5 BA
⇒
⇒

∴
Hence, the required vector is

Q.5. Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.
Ans.
Let the given points are A( k , - 10 , 3), B(1,- 1, 3) and C(3, 5, 3)

If A, B and C are collinear, then

Squaring both sides, we have
√

= 9 + k²- 6k + 225
⇒
Dividing by 2, we get
⇒
⇒

⇒(Dividing by 2)
Squaring both sides, we get
⇒ 10(k² – 2k + 82) = 784 + k² – 56k
⇒ 10k² – 20k + 820 = 784 + k² – 56k
⇒ 10k² – k² – 20k + 56k + 820 – 784 = 0
⇒ 9k² + 36k + 36 = 0
⇒ k² + 4k + 4 = 0
⇒ (k + 2)² = 0
⇒ k + 2 = 0
⇒ k = – 2
Hence, the required value is k = – 2

Q.6. A vectoris inclined at equal angles to the three axes. If the magnitude of  is 2 √3 units, find
Ans.
Since, the vectormakes equal angles with the axes, their direction cosines should be same
∴ l = m = n
We know that
l² + m² + n² = 1
⇒ l² + l² + l² = 1
⇒
∴
We know that

Hence, the required value of

Q.7. A vector has magnitude 14 and direction ratios 2, 3, – 6. Find the direction cosines and components of  , given that makes an acute angle with x-axis.
Ans.
Letbe three vectors such thatand
If l, m and n are the direction cosines of vector , then


We know that l² + m² + n² = 1
∴

∴ k = ± 2 and l =

∴

⇒
Hence, the required direction cosines areand the components of


Q.8. Find a vector of magnitude 6, which is perpendicular to both the vectors and
Ans.
Letand
We know that unit vector perpendicular to



Now the vector of magnitude 6 =

Hence, the required vector is

Q.9. Find the angle between the vectors.
Ans.
Let and let θ be the angle between
∴

∴
Hence, the required value of θ is

Q.10. If show that   Interpret the result geometrically?
Ans.
Given that
So,
⇒
⇒
⇒
⇒...(i)
Now
⇒
⇒
⇒
⇒
∴...(ii)
From eq. (i) and (ii) we get

Hence proved.
Geometrical Interpretation
According to figure, we have
Area of parallelogram ABCD is

Since, the parallelograms on the same base and between the same parallel lines are equal in area


Q.11. Find the sine of the angle between the vectors and
Ans.
Given that
We know that
∴





∴
⇒
Hence,

Q.12. If A, B, C, D are the points with position vectors
respectively, find the projection of
Ans.
Here, Position vector of A =
Position vector of B =
Position vector of C =
Position vector of D =

Projection of

Hence, the required projection = √21 .

Q.13. Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).
Ans.
Given that A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1)

Area of ΔABC ==
=

Hence, the required area is

Q.14. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
Ans.
Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.
Let
∴ Area of parallelogram ABCD =
Now Area of parallelogram ABFE =



Hence proved.

LONG ANSWER TYPE QUESTIONS

Q.15. Prove that in any triangle ABC,where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
Ans.
Here, in the given figure, the components of c are c cos A and c sin A.
∴
In DBDC,
a² = CD² + BD²
⇒ a² = (b – c cos A)² + (c sin A)²
⇒ a² = b² + c² cos² A – 2bc cos A + c² sin² A
⇒ a² = b² + c²(cos² A + sin² A) – 2bc cos A
⇒ a² = b² + c² – 2bc cos A
⇒ 2bc cos A = b² + c² - a²
∴
Hence Proved.

Q.16. Ifdetermine the vertices of a triangle, show that
gives the vector area of the triangle. Hence deduce the condition that the three points are collinear. Also find the unit vector normal to the plane of the triangle.
Ans.
Since,are the vertices of ΔABC




For three vectors are collinear, area of ΔABC = 0

which is the condition of collinearity of
Letbe th e unit vector normal to the plane of the ΔABC
∴
⇒

Q.17. Show that area of the parallelogram whose diagonals are given by is. Also find the area of the parallelogram whose diagonals are and
Ans.
Let ABCD be a parallelogram such that,

∴ by law of triangle, we get

Adding eq. (i) and (ii) we get,

Subtracting eq. (ii) from eq. (i) we get




So, the area of the parallelogram ABCD =
Now area of parallelogram whose diagonals areand



Hence, the required area is

Q.18. Iffind a vector  such that  
Ans.
Let
Also given that
Since,
∴

On comparing the like terms, we get
c₃ – c₂ = 0 ...(i)
c₁ – c₃ = 1 ...(ii)

and c₂ – c₁ = –1 ...(iii)
Now

∴ c₁ + c₂ + c₃ = 3 ...(iv)
Adding eq. (ii) and eq. (iii) we get,
c₂ – c₃ = 0 ...(v)
From (iv) and (v) we get
c₁ + 2c₂ = 3 ...(vi)
From (iii) and (vi) we get
Adding

∴
c₃ – c₂ = 0
⇒
∴
Now c₂ – c₁ = – 1 ⇒
⇒
∴
Hence,

OBJECTIVE TYPE QUESTIONS

Q.19. The vector in the direction of the vectorthat has magnitude 9 is
(a)
(b)
(c)
(d)
Ans. (c)
Solution.
Let
Unit vector in the direction of

∴ Vector of magnitude 9 =
Hence, the correct option is (c).

Q.20. The position vector of the point which divides the join of points
in the ratio 3 : 1 is
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
The given vectors areand the ratio is 3 : 1.
∴ The position vector of the required point c which divides the join of the given vectors



Hence, the correct option is (d).

Q.21. The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is
(a)
(b)
(c)
(d)
Ans. (c)
Solution.
Let A and B be two points whose coordinates are given as (2, 5, 0) and (– 3, 7, 4)
∴
⇒
Hence, the correct option is (c).

Q.22. The angle between two vectorswith magnitudes √3 and 4, respectively, and
(a) π/6
(b) π/3
(c) π/2
(d) 5π/2
Ans. (b)
Solution.
Here, given that
∴ From scalar product, we know that

⇒
⇒
∴
Hence, the correct option is (b).

Q.23. Find the value of λ such that the vectors
are orthogonal
(a) 0 
(b) 1 
(c) 3/2
(d) -5/2
Ans. (d)
Solution.
Given that
Sinceare orthogonal  
∴

⇒ 2 + 2λ + 3 = 0
⇒ 5 + 2λ = 0 ⇒
Hence, the correct option is (d).

Q.24. The value of λ for which the vectors
are parallel is
(a) 2/3
(b) 2/3
(c) 5/2
(d) 2/5
Ans. (a)
Solution.
Let

Since the given vectors are parallel,
∴ Angle between them is 0°
so


Squaring both sides, we get
900 + λ² + 60λ = 46(20 + λ²)
⇒ 900 + λ² + 60λ = 920 + 46λ²
⇒ λ² – 46λ² + 60λ + 900 – 920 = 0
⇒ - 45λ² + 60λ - 20 = 0
⇒ 9λ² – 12λ + 4 = 0
⇒(3λ – 2)² = 0
⇒3λ – 2 = 0
⇒ 3λ = 2
∴ λ = 2/3
Alternate method:
Let
If
∴
⇒
Hence, the correct option is (a).

Q.25. The vectors from origin to the points A and B are
,respectively, then the area of triangle OAB is   
(a) 340 
(b) √25 
(c) √229 
(d)
Ans. (d)
Solution.
Let O be the origin
∴
and
∴ Area of ΔOAB =

Hence the correct option is (d).

Q.26. For any vectorthe value ofis equal to
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
Let
∴
Now,


∴
Similarly
and

Hence, the correct option is (d).

Q.27. Ifthen value ofis
(a) 5 
(b) 10 
(c) 14 
(d) 16
Ans. (d)
Solution.
Given that
∴

⇒ 12 = 10 × 2 × cos θ
⇒
∴
⇒
⇒
Now

Hence, the correct option is (d).

Q.28. The vectorsare coplanar if
(a) λ = –2 
(b) λ = 0 
(c) λ = 1 
(d) λ = – 1
Ans. (a)
Solution.
Let

Ifare coplanar, then

∴

⇒ l(λ² – 1) – 1 (λ + 2) + 2(–1 – 2l) = 0
⇒ λ³ – λ – λ – 2 – 2 – 4λ = 0
⇒ λ³ – 6λ – 4 = 0
⇒ (λ + 2) (λ² – 2λ – 2) = 0
⇒ λ = – 2 or λ² – 2λ – 2 = 0
⇒
⇒
∴
Hence, the correct option is (a).

Q.29. Ifare unit vectors such thatthen the value of

(a) 1
(b) 3
(c) -3/2
(d) None of these
Ans. (c)
Solution.
Given that
and
∴

⇒
⇒
⇒
⇒
Hence, the correct option is (c).

Q.30. Projection vector ofis
(a)
(b)
(c)
(d)
Ans. (a)
Solution.
The projection vector of 
Hence, the correct option is (a).

Q.31. Ifare three vectors such that
then value of
(a) 0 
(b) 1 
(c) – 19 
(d) 38
Ans. (c)
Solution.
Given that    
and


⇒
⇒
⇒
⇒
⇒
⇒
Hence, the correct option is (c).

Q.32. Ifand −3 ≤ λ ≤ 2 , then the range ofis   
(a) [0, 8]
(b) [– 12, 8]     
(c) [0, 12]     
(d) [8, 12]
Ans. (b)
Solution.
Given that
Now
Here - 3 ≤ λ ≤ 2
⇒ - 3.4 ≤ 4λ  ≤ 2.4
⇒ - 12 ≤ 4λ  ≤ 8
∴ 4λ  = [- 12, 8]
Hence, the correct option is (b).

Q.33. The number of vectors of unit length perpendicular to the vectorsand
(a) one 
(b) two 
(c) three 
(d) infinite
Ans. (b)
Solution.
The number of vectors of unit length perpendicular to vectors

∴
So, there will be two vectors of unit length perpendicular to vectors
Hence, the correct option is (b).

FILL IN THE BLANKS

Q.34. The vectorbisects the angle between the non-collinear vectorsif ________.
Ans.
If vector bisects the angle between non-collinear vectors 
then the angle betweenis equal to the angle between
So,  ...(i)

Also,[∵ θ is same]   ...(ii)
From eq. (i) and eq. (ii) we get,

⇒
⇒
Hence, the required filler is

Q.35. Iffor some non-zero vectorthen the value of is   ________
Ans.
Ifis a non-zero vector, thencan be in the same plane.
Since angles between  and    are zero i.e. θ = 0

Hence the required value is 0.

Q.36. The vectorsa re the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.
Ans.
Given that
and

Let θ be the angle between the two diagonal vectors
then



Hence the value of required filler is

Q.37. The values of k for which is parallel toholds true are _______.
Ans.
Given that

Now sinceis parallel to
Here we see that atbecome null vector and then it will not be
parallel to
Now sinceis parallel to
Here we see that atbecome null vector and then it will not be
parallel to 

Hence, the required value of k ∈ (- 1, 1) and k ≠

Q.38. The value of the expressionis _______.
Ans.


Hence, the value of the filler is

Q.39. Ifandis equal to _______.
Ans.

⇒
⇒
⇒
⇒
⇒
⇒
∴
Hence, the value of the filler is 3.

Q.40. Ifany non-zero vector, then
equals _______.
Ans.
Let
∴
= a₁


Hence, the value of the filler is

State True or False in each of the following Exercises.
Q.41. Ifthen necessarily it implies
Ans.
Ifthenwhich is true.
Hence, the statement is True.

Q.42. Position vector of a point P is a vector whose initial point is origin.
Ans.
True

Q.43. Ifthen the vectorsare orthogonal.
Ans.
Given that
Squaring both sides, we get

⇒
⇒
⇒
which implies thatare orthogonal.
Hence the given statement is True.

Q.44. The formulais valid for non-zero vectors

Ans.

Hence, the given statement is False.

Q.45. Ifare adjacent sides of a rhombus, then
Ans.
If
So the angle between the adjacent sides of the rhombus should be 90° which is not possible.
Hence, the given statement is False.