Class 12 RD Sharma Solutions – Chapter 24 Scalar or Dot Product – Exercise 24.1 | Set 2
Question 17. If and , then express in the form of where is parallel to and is perpendicular to .
Solution:
Given,
According to question
also = 0
Now,
⇒
⇒
⇒
Now,
⇒
⇒ 3(2-3λ)+4(1-4λ)-5(4+5λ) = 0
⇒ 6-9λ+4-16λ-20-25λ = 0
⇒ -10 -50λ = 0
⇒ λ = -1/5
Question 18. If either or , then . But, The converse need not be true. Justify your answer with an example.
Solution:
Given,
or then
Suppose
But,
= √(2)2+(1)2+(1)2
= √4+1+1
= √6 ≠ 0
= √(1)2+(1)2+(1)2
= √3 ≠ 0
Hence Proved
Question 19. Show that the vectors form a right-angled triangle.
Solution:
Given,
To prove given vectors form a right angle triangle
= √(32+(-2)2+12) = √14
= √(12+(-3)2+52) = √35
= √(22+12+(-4)2) = √21
= 14 + 21 = 35
Since, (Pythagoras Theorem)
Hence, and form a right angled triangle.
Question 20. If , and are such that is perpendicular to , then find the value of λ.
Solution:
Given:
Now,
⇒
⇒
⇒ (2 – λ)3 + (2 + 2λ) + 0 = 0
⇒ 6 – 3λ + 2 + 2λ =0
⇒ λ = 8
Question 21. Find the angles of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4), and C(5, 7, 1).
Solution:
Given that angle of a triangle whose vertices are A (0, -1. -2), B (3, 1, 4) and C(5, 7, 1).
= √98 = 7√2
Now,
= (3 × 2 + 2 × 6 – 6 × 3) = 0
Thus, we can say AB is perpendicular to BC.
Hence, AB = BC = 7, ∠A =∠C and ∠B = 90°
∠A + ∠B + ∠C = 180°
2∠A = 180° – 90°
∠A = 45°
∠C = 45°
∠B = 90°
Question 22. Find the magnitude of two vectors and , having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.
Solution:
We know
⇒ 1/ 2 =
⇒ 1/2 = (1/2)
⇒
or
⇒
Question 23. Show that the points whose position vector are form a right triangle.
Solution:
Given that positions vectors
Now,
⇒
⇒
⇒
Now,
= 2 – 3 – 20 = -21
= -3 – 6 – 5 = -14
= -6 + 2 + 4 = 0
So, AB is perpendicular to CA or the given position vectors form a right-angled triangle.
Question 24. If the vertices A, B, C of △ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, what is the magnitude of ∠ABC?
Solution:
Given the vertices of △ABC are A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2)
Now,
=
Or,
We know that
(2 × 1) + (2 × 1) + (3 × 2)
= 2 + 2 + 6 = 10
Now, = √17
= √6
Therefore,
cos θ =
cos θ = 10/ √(17×6)
θ = cos-1(10/√102)
Question 25. If A, B, C have position vectors (0, 1, 1), (3, 1, 5), (0, 3, 3) respectively, show that △ABC is right-angled at C.
Solution:
Given, position vectors A(0, 1, 1), B(3, 1, 5), C(0, 3, 3)
Now,
=
= 2 × 2 – 2 × 2 = 0
Thus, and are perpendicular hence △ABC is right-angled at C
Question 26. Find the projection of on , where and.
Solution:
Given:
To find the projection of on
Now, Projection of =
=
= 6/9 × 3
= 2
Question 27. If and , then show that the vectors and are orthogonal.
Solution:
Given:
To prove
Taking LHS
=
=
= √35 – √35
= 0
Thus, the given vectors and are orthogonal.
Question 28. A unit vector makes angle π/2 and π/3 with and respectively and an acute angle θ with . Find the angle θ and components of .
Solution:
Let us assume
We know that
a12+ a22 + a32 = 1 ….(1)
So,
(1)(1)(1/√2) = a1
a1 = 1/√2
Again we take
(1)(1)(1/2) = a2
a2 = 1/2
Put all these values in eq(1) to find the value of a3
(1/√2)2 + (1/2)2 + a32 = 1 ….(1)
a32 = 1/4
a3 = 1/2
Now we find the value of θ
(1)(1)cosθ = 1/2
cosθ = 1/2
cosθ = π/3
and components of
Question 29. If two vectors and are such that = 2, = 1, and =1. Find the value of
Solution:
Given,
=
=
= 6(2)2 + 11(1) – 35(1)2
= 24 + 11 – 35
= 35 – 35 = 0
Question 30. If is a unit vector, then find in each of the following:
(i)
Solution:
Given,
⇒
⇒
⇒
⇒
⇒
(ii)
Solution:
Given,
⇒
⇒
⇒
⇒
⇒ =√13
Question 31. Find and , if
(i) = 12 and
Solution:
Given, = 12
⇒
⇒ = 12
⇒ = 12
⇒ = 12
⇒ = 2
So,
= 4
(ii) = 8 and = 8
Solution:
Given, = 8
⇒
⇒
⇒
⇒
⇒ = √(8/63)
So,
= 8√(8/63)
(iii)= 3 and = 2
Solution:
Given,
⇒
⇒
⇒
⇒ 3= 3
⇒ = 1
So,
= 2
Question 32. Find , if
(i) and
Solution:
We have,
⇒
⇒= 22 – 2 × 8 + 52
⇒ = 4 – 16 + 25
⇒ = 13
⇒= √13
(ii) = 3, = 4 and = 1
Solution:
We have,
⇒
⇒ = 32 – 2 × 1 + 42
⇒ = 9 – 2 + 16
⇒ = 23
⇒ = √23
(iii) and = 4
Solution:
We have,
⇒
⇒= 22 – 2 × 4 + 32
⇒ = 4 – 8 + 9
⇒ = 5
⇒ = √5
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