Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 3
Question 33. Find the angle between the two vectors and , if
(i) =√3, = 2 and = √6
Solution:
We know,
⇒ √6 = 2√3 cos θ
⇒ cos θ = 1/√2
⇒ θ = cos-1(1/√2)
⇒ θ = π/4
(ii) = 3, = 3 and = 1
Solution:
We know,
⇒ 1 = 3×3 cos θ
⇒ cos θ = 1/9
⇒ θ = cos-1(1/9)
Question 34. Express the vector as the sum of two vectors such that one is parallel to the vector and other is perpendicular to
Solution:
Given,
Let the two vectors be
Now, ….(1)
Assuming is parallel to
Then, ……(2)
is perpendicular to
Then, ……(3)
From eq(1)
⇒
⇒
⇒
From eq(3)
⇒
⇒ (5-3λ)3+(5-λ)=0
⇒ 15-9λ+5-λ=0
⇒ -10λ = -20
⇒ λ=2
From eq(2)
Question 35. If and are two vectors of the same magnitude inclined at an angle of 30° such that = 3, find
Solution:
Given that two vectors of the same magnitude inclined at an angle of 30°, and
To find
We know,
⇒ 3 =
⇒ 3 =
⇒ 3 = (√3/2)
⇒= 6/√3
⇒
Question 36. Express as the sum of a vector parallel and a vector perpendicular to
Solution:
Assuming
Let the two vectors be
Now,
or ….(1)
Assumingis parallel to
then, …(2)
is perpendicular to
then,……(3)
Putting eq(2) in eq(1), we get
⇒
⇒
⇒
From eq(3)
⇒
⇒ (2 – 2λ)2 – (1 + 4λ)4 – (3 + 2λ)2 = 0
⇒ 4 – 4λ – 4 – 16λ – 6 – 4λ = 0
⇒ 24λ = -6
⇒ λ = -6/24
From eq(2)
Question 37. Decompose the vector into vectors which are parallel and perpendicular to the vector
Solution:
Let and
Let be a vector parallel to
Therefore,
to be decomposed into two vectors
⇒
⇒
Now, is perpendicular to
or
⇒
⇒ 6 – λ – 3 – λ – 6 – λ = 0
⇒ λ = -1
Therefore, the required vectors are and
Question 38. Let and . Find λ such that is orthogonal to
Solution:
Given,
According to question
⇒
⇒
⇒
⇒ 25 + 1 + 49 = 1 + 1 + λ2
⇒ λ2 = 73
⇒ λ = √73
Question 39. If and , what can you conclude about the vector ?
Solution:
Given, ,
Now,
We conclude that or or θ = 90°
Thus, can be any arbitrary vector.
Question 40. If is perpendicular to both and , then prove that it is perpendicular to both and
Solution:
Given is perpendicular to both and
….(1)
….(2)
To prove and
Now,
⇒ [From eq(1) and (2)]
Again,
⇒ [From eq(1) and (2)]
Hence Proved
Question 41. If and , prove that
Solution:
Given, and
To prove
Taking LHS
=
=
=
Taking RHS
=
=
LHS = RHS
Hence Proved
Question 42. If are three non- coplanar vectors such that then show that is the null vector.
Solution:
Given that
So either or
Similarly,
Either or
Also,
So or
But can’t be perpendicular to and because are non-coplanar.
So = 0 oris a null vector
Question 43. If a vector is perpendicular to two non- collinear vectors and , then is perpendicular to every vector in the plane of and
Solution:
Given that is perpendicular to and
Let be any vector in the plane of and and is the linear combination of and
[x, y are scalars]
Now
⇒
⇒
⇒
⇒
Therefore, is perpendicular to i.e. is perpendicular to every vector.
Question 44. If , how that the angle θ between the vectors and is given by cos θ =
Solution:
Given that
⇒
⇒
⇒
⇒
⇒
⇒
⇒ cos θ =
Question 45. Let and be vector such . = 3, = 4 and = 5, then find
Solution:
Given that and are vectors such that . = 3, = 4 and =5,
To find
Taking
Squaring on both side, we get
⇒
⇒
⇒
⇒
⇒
Therefore,
Question 46. Let and be three vectors. Find the values of x for which the angle between and is acute and the angle between and is obtuse.
Solution:
Given
Case I: When angle between and is acute:-
>0
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⇒ x2 – 2 – 2 > 0
⇒ x2 > 4
x ∈ (2, -2)
Case II: When angle between and is obtuse:-
⇒
⇒ x2 – 5 – 4 < 0
⇒ x2 < 9
x ∈ (3, -3)
Therefore, x ∈ (-3, -2)∪(2, 3)
Question 47. Find the value of x and y if the vectorsand are mutually perpendicular vectors of equal magnitude.
Solution:
Given are mutually perpendicular vectors of equal magnitude.
⇒ 32 + x2 + (-1)2 = 22 + 12 + y2
⇒ x2+10 = y2+5
⇒ x2 – y2 + 5 = 0 ….(1)
Now,
⇒ 6 + x – y = 0
⇒ y = x + 6 …..(2)
From eq(1)
x2 – (x + 6)2 + 5 = 0
⇒ x2 – (x2 + 36 – 12x) + 5 = 0
⇒ -12x – 31 = 0
⇒ x = -31/12
Now, y = -31/12 + 6
y = 41/12
Question 48. If and are two non-coplanar unit vectors such that , find
Solution:
Given that and are two non-coplanar unit vectors such that
To find
Now,
Now,
=
= 6 – 13(1/2) – 5
= 1 – 13/2
= -11/2
Question 49. If are two vectors such that || = , then prove that is perpendicular to
Solution:
To prove
Now,
Squaring on both side, we get
⇒
⇒
⇒
⇒
Therefore, is perpendicular to
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