Notes of vectors

1. 𝜃
𝑃 𝑄
𝑅
𝑎 × 𝑏
𝑎
𝑏

2. Plane and its equation in Vector Form
Properties of vector product
Vector product of two vectors

4. Class Description:
BHLP Session:
Timings:
Duration:
Video Number:
Vectors
10
0:03:55-0:10:45
0:06:50
01

5. Find a line passing through 𝑎 and perpendicular to both the lines
𝐿1 ∶ 𝑚 + 𝜆𝑞, 𝐿2 ∶ 𝑛 + 𝜇𝑝
𝐿1
𝑎
𝐿2
∴ Equation of the required line is 𝑟 = 𝑎 + 𝑡 𝑝 × 𝑞
The vector parallel to the line 𝐿1: 𝑞
The vector parallel to the line 𝐿2: 𝑝
The vector perpendicular to the line 𝐿1
and 𝐿2: 𝑝 × 𝑞
Where 𝑎 is the position vector of fixed point,
and 𝑝 × 𝑞 is the vector parallel to the
required line.
𝑝 × 𝑞
Solution:

6. Vector (or Cross) Product of Two Vectors
𝑏
𝑎
𝑎 × 𝑏
𝑎 × 𝑏 ⋅ 𝑎 = 0
𝑎 × 𝑏 ⋅ 𝑏 = 0
𝑚 × 𝑟 ⋅ 𝑚 = 0
𝑎 × 𝑏 × 𝑐 ⋅ 𝑎 × 𝑏 = 0

7. If 𝑎 = 1, 𝑏 = 1, 𝑐 = 2 and 𝑎 × 𝑎 × 𝑐 − 𝑏 = 0, then find angle
between 𝑎 and 𝑐.

8. If 𝑎 = 1, 𝑏 = 1, 𝑐 = 2 and 𝑎 × 𝑎 × 𝑐 − 𝑏 = 0, then find angle
between 𝑎 and 𝑐.
Solution:

9. 𝑎 × 𝑖 ⋅ 𝑖 + 𝑎 × 𝑗 ⋅ 𝑗 + 𝑎 × 𝑘 ⋅ 𝑘 =?
−1
0
1
Cannot be determined

10. −1
0
1
Cannot be determined
𝑎 × 𝑖 ⋅ 𝑖 + 𝑎 × 𝑗 ⋅ 𝑗 + 𝑎 × 𝑘 ⋅ 𝑘 =?

11. 𝑎 × 𝑖 ⋅ 𝑖 + 𝑎 × 𝑗 ⋅ 𝑗 + 𝑎 × 𝑘 ⋅ 𝑘 =?
Solution:

12. 𝑎 × 𝑖 2
+ 𝑎 × 𝑗 2
+ 𝑎 × 𝑘
2
= ?

13. 𝑎 × 𝑖 2
+ 𝑎 × 𝑗 2
+ 𝑎 × 𝑘
2
= ?

15. Class Description:
BHLP Session:
Timings:
Duration:
Video Number:
Vectors
10
0:19:48-0:26:47
0:06:59
02

16. If 𝑎 + 𝑏 + 𝑐 = 0, then prove that 𝑎 × 𝑏 = 𝑏 × 𝑐 = 𝑐 × 𝑎
𝑎 + 𝑏 + 𝑐 = 0 ⋯ 𝑖
𝑎 × 𝑎 + 𝑏 × 𝑎 + 𝑐 × 𝑎 = 0 × 𝑎
⇒ 𝑐 × 𝑎 = −𝑏 × 𝑎
⇒ 𝑐 × 𝑎 = 𝑎 × 𝑏 ⋯ (𝑖𝑖)
𝑖 × 𝑏 ⇒ 𝑎 × 𝑏 + 𝑐 × 𝑏 = 0 × 𝑏
⇒ 𝑎 × 𝑏 = −𝑐 × 𝑏
⇒ 𝑎 × 𝑏 = 𝑏 × 𝑐 ⋯ (𝑖𝑖𝑖)
∴ 𝑎 × 𝑏 = 𝑏 × 𝑐 = 𝑐 × 𝑎
From (𝑖𝑖) and (𝑖𝑖𝑖),
Solution:

17. Vector (or Cross) Product of Two Vectors
If 𝑎 × 𝑏 = 𝑐, 𝑏 × 𝑐 = 𝑎, 𝑐 × 𝑎 = 𝑏 (given), then
𝑎 × 𝑏 = 𝑐 ⇒ 𝑐 ⊥ 𝑎, 𝑐 ⊥ 𝑏
𝑏 × 𝑐 = 𝑎 ⇒ 𝑎 ⊥ 𝑏, 𝑎 ⊥ 𝑐
𝑎 ⊥ 𝑏 ⊥ 𝑐
Now, 𝑎 × 𝑏 = 𝑐 ⇒ 𝑎 × 𝑏 = 𝑐
⇒ 𝑎 𝑏 sin 90° = 𝑐 ⇒ 𝑎 𝑏 = 𝑐 ⋯ (𝑖)
Similarly, 𝑏 𝑐 = 𝑎 ⋯ (𝑖𝑖)

18. Similarly, 𝑐 = 𝑎 = 1
∴ 𝑎 = 𝑏 = 𝑐 = 1
∴ 𝑎 = 𝑖, 𝑏 = 𝑗, 𝑐 = 𝑘
Three orthogonal unit vectors
𝑎 𝑏
𝑏 𝑐
=
𝑐
𝑎
⇒ 𝑎 = 𝑐
From i , (𝑖𝑖),
∴ 𝑏 = 1 From i

19. If 𝑎, b and 𝑐 are 3 non-collinear vectors such that 𝑎 × 𝑏 = 𝑐, 𝑏 × 𝑐 = 𝑎, 𝑐 ×
𝑎 = 𝑏, then 𝑎 + 𝑏 + c = ?
𝟑
𝟏
𝟑
𝟓

20. If 𝑎, b and 𝑐 are 3 non-collinear vectors such that 𝑎 × 𝑏 = 𝑐, 𝑏 × 𝑐 = 𝑎, 𝑐 ×
𝑎 = 𝑏, then 𝑎 + 𝑏 + c = ?
𝟑
𝟏
𝟑
𝟓

21. 𝑖. If 𝑎 × 𝑏 = c × 𝑑 and 𝑎 × 𝑐 = 𝑏 × 𝑑, then 𝑎 − 𝑑 × 𝑏 − 𝑐 = ?
𝑖𝑖. If 𝑎 = 𝑖 + 2j, 𝑏 = 3j − 5k and 𝑟 × 𝑎 = 𝑎 × 𝑏, 𝑟 × 𝑏 = 𝑏 × 𝑎, then find the unit
vector in direction of 𝑟.

22. Solution:
𝑖. If 𝑎 × 𝑏 = c × 𝑑 and 𝑎 × 𝑐 = 𝑏 × 𝑑, then 𝑎 − 𝑑 × 𝑏 − 𝑐 = ?
𝑖𝑖. If 𝑎 = 𝑖 + 2j, 𝑏 = 3j − 5k and 𝑟 × 𝑎 = 𝑎 × 𝑏, 𝑟 × 𝑏 = 𝑏 × 𝑎, then find the unit
vector in direction of 𝑟.

23. Solution:
𝑖. If 𝑎 × 𝑏 = c × 𝑑 and 𝑎 × 𝑐 = 𝑏 × 𝑑, then 𝑎 − 𝑑 × 𝑏 − 𝑐 = ?
𝑖𝑖. If 𝑎 = 𝑖 + 2j, 𝑏 = 3j − 5k and 𝑟 × 𝑎 = 𝑎 × 𝑏, 𝑟 × 𝑏 = 𝑏 × 𝑎, then find the unit
vector in direction of 𝑟.

24. For 2 particular vectors 𝐴, 𝐵 it is known that 𝐴 × 𝐵 = 𝐵 × 𝐴,
one of the two vectors is a multiple of the other.[True/False]

25. Solution:
For 2 particular vectors 𝐴, 𝐵 it is known that 𝐴 × 𝐵 = 𝐵 × 𝐴,
one of the two vectors is a multiple of the other.[True/False]

26. Let 𝑎, 𝑏 and 𝑐, 𝑑 each determine a plane 𝑃1 and 𝑃2 respectively,
such that 𝑎 × 𝑏 × 𝑐 × 𝑑 = 0. Find the angle between 𝑃1 and 𝑃2.

27. Solution:
Let 𝑎, 𝑏 and 𝑐, 𝑑 each determine a plane 𝑃1 and 𝑃2 respectively,
such that 𝑎 × 𝑏 × 𝑐 × 𝑑 = 0. Find the angle between 𝑃1 and 𝑃2.

28. ′𝑃′ is a point inside the ∆𝐴𝐵𝐶, such that 𝐵𝐶 𝑃𝐴 + 𝐶𝐴 𝑃𝐵 + 𝐴𝐵 𝑃𝐶 = 0,
then for the triangle ∆𝐴𝐵𝐶 the point 𝑃 is:

29. Solution:
′𝑃′ is a point inside the ∆𝐴𝐵𝐶, such that 𝐵𝐶 𝑃𝐴 + 𝐶𝐴 𝑃𝐵 + 𝐴𝐵 𝑃𝐶 = 0,
then for the triangle ∆𝐴𝐵𝐶 the point 𝑃 is:

30. …

31. Class Description:
BHLP Session:
Timings:
Duration:
Video Number:
12th Vectors
10
0:42:07-0:46:30
0:04:23
03

32. ⇒ 𝑝 × 𝑞 − 𝑝 × 𝑟 = 0
If 𝑝 ≠ 0, 𝑝 ⋅ 𝑞 = 𝑝 ⋅ 𝑟 & 𝑝 × 𝑞 = 𝑝 × 𝑟. What’s the relation between 𝑞 and 𝑟 ?
⇒ 𝑝 × 𝑞 − 𝑟 = 0
⇒ 𝑝 = 0 𝑜𝑟 𝑞 − 𝑟 = 0 𝑜𝑟 𝑝 ∥ 𝑞 − 𝑟
⇒ 𝑝 ⋅ 𝑞 − 𝑝 ⋅ 𝑟 = 0
⇒ 𝑝 ⋅ 𝑞 − 𝑟 = 0
⇒ 𝑝 = 0 𝑜𝑟 𝑞 − 𝑟 = 0 𝑜𝑟 𝑝 ⊥ 𝑞 − 𝑟
𝑝 ⋅ 𝑞 = 𝑝 ⋅ 𝑟 𝑝 × 𝑞 = 𝑝 × 𝑟
Case 1: Case 2:
From case 1 and case 2, we get, 𝑝 = 0 or 𝑞 − 𝑟 = 0
⇒ 𝑞 = 𝑟
But 𝑝 ≠ 0 ( Given )
∴ 𝑞 − 𝑟 = 0
Solution:

33. ⇒ 𝑢 − 𝑎 × 𝑏 = 0
If 𝑢 × 𝑏 = 𝑎 × 𝑏 and 𝑢 ⋅ 𝑐 = 0 then 𝑢 =?
⇒ 𝑢 − 𝑎 ∥ 𝑏 ⇒ 𝑢 − 𝑎 = 𝜆𝑏
Given, 𝑢 × 𝑏 = 𝑎 × 𝑏
⇒ 𝑢 × 𝑏 − 𝑎 × 𝑏 = 0
⇒ 𝑢 = 𝑎 + 𝜆𝑏 … (𝑖)
⇒ 𝑎 + 𝜆𝑏 ⋅ 𝑐 = 0
⇒ 𝑎 ⋅ 𝑐 = −𝜆𝑏 ⋅ 𝑐 ⇒ 𝜆 =
−𝑎 ⋅ 𝑐
𝑏 ⋅ 𝑐
𝑢 = 𝑎 −
𝑎 ⋅ 𝑐
𝑏 ⋅ 𝑐
𝑏
𝑢 ⋅ 𝑐 = 0 Given
Substituting the value of 𝜆 in equation 𝑖 ,
From (𝑖)
Solution:

34. Let 𝑢 = 𝑖 + 𝑗, 𝑣 = 𝑖 − 𝑗, 𝑤 = 𝑖 + 3𝑗 + 5𝑘 and 𝑢 ⋅ 𝑛 = 𝑣 ⋅ 𝑛 = 0.
Then find 𝑤 ⋅ 𝑛 .

35. Solution:
Let 𝑢 = 𝑖 + 𝑗, 𝑣 = 𝑖 − 𝑗, 𝑤 = 𝑖 + 3𝑗 + 5𝑘 and 𝑢 ⋅ 𝑛 = 𝑣 ⋅ 𝑛 = 0.
Then find 𝑤 ⋅ 𝑛 .

37. Class Description:
BHLP Session:
Timings:
Duration:
Video Number:
12th Vectors
10
0:52:09-0:57:56
0:05:47
04

38. 𝑢 × 𝑣 = 𝑎 + 𝑏 × 𝑎 − 𝑏 = 0 − 𝑎 × 𝑏 + 𝑏 × 𝑎 − 0
If 𝑢 = 𝑎 + 𝑏 and 𝑣 = 𝑎 − 𝑏 and 𝑎 = 𝑏 = 2, then 𝑢 × 𝑣 =?
= 2 𝑏 × 𝑎
𝑎 = 𝑏 = 2
= −𝑎 × 𝑏 + 𝑏 × 𝑎
= 𝑏 × 𝑎 + 𝑏 × 𝑎
⇒ 𝑢 × 𝑣 = 2 𝑏 × 𝑎
⇒ 𝑢 × 𝑣 2
= 4 𝑏 × 𝑎
2
⇒ 𝑢 × 𝑣 2
= 4 𝑎 2
𝑏
2
− 𝑎 ⋅ 𝑏
2
⇒ 𝑢 × 𝑣 = 4 16 − 𝑎 ⋅ 𝑏
2
⇒ 𝑢 × 𝑣 = 2 16 − 𝑎 ⋅ 𝑏
2

39. If 𝑎, 𝑏 and 𝑐 such that 𝑎 + 𝑏 + 𝑐 = 1, 𝑐 = 𝑟 𝑎 × 𝑏 , 𝑎 =
1
2
, 𝑏 =
1
3
,
𝑐 =
1
6
then find angle between 𝑎 and 𝑏.
⇒
1
2
+
1
3
+
1
6
+ 2 𝑎 ⋅ 𝑏 + 𝑏 ⋅ 𝑐 + 𝑐 ⋅ 𝑎 = 1
⇒ 𝑎 ⋅ 𝑏 + 𝑏 ⋅ 𝑐 + 𝑐 ⋅ 𝑎 = 0 ⋯ 𝑖
𝑎 + 𝑏 + 𝑐 = 1
⇒ 𝑎 2
+ 𝑏
2
+ 𝑐 2
+ 2 𝑎 ⋅ 𝑏 + 𝑏 ⋅ 𝑐 + 𝑐 ⋅ 𝑎 = 12
Given: 𝑐 = 𝑟 𝑎 × 𝑏 ⇒ 𝑐 ∥ 𝑎 × 𝑏
⇒ 𝑐 ⊥ 𝑎 and 𝑐 ⊥ 𝑏
⇒ 𝑐 ⋅ 𝑎 = 𝑐 ⋅ 𝑏 = 0 ⋯ (𝑖𝑖)
From (𝑖) and 𝑖𝑖 ,
𝑎 ⋅ 𝑏 = 0⇒ 𝑎 ⊥ 𝑏 = 0
⇒ Angle =
𝜋
2

40. If 𝑎 = 𝑏 = 1 and 𝑎 + 𝑏 = 3, 𝑐 is a vector such that 𝑐 − 𝑎 − 2𝑏 = 3 𝑎 × 𝑏
then 𝑐 ⋅ 𝑏 =?

41. Let 𝑎 = 𝑖 − 𝑗 , 𝑏 = 𝑖 + 𝑗 + 𝑘 and 𝑐 be vector such that 𝑎 × 𝑐 + 𝑏 = 0
and 𝑎 ⋅ 𝑐 = 4, then 𝑐 2
is equal to:
JEE Main Jan 2019

42. …

43. Class Description:
BHLP Session:
Timings:
Duration:
Video Number:
12th Vectors
10
1:02:41-1:08:18
0:05:37
05

44. Area of Parallelogram
𝑎
𝑏
𝜃
𝑏 sin 𝜃
Area of parallelogram = (
1
2
𝑎 𝑏 sin 𝜃) × 2
Area of parallelogram = 𝑎 × 𝑏
Area of parallelogram = 𝑑1 × 𝑑2
𝑑1
𝑑2 𝜃
Geometrically, 𝑎 × 𝑏 represents vector area of a parallelogram having adjacent sides as
𝑎 and 𝑏.

45. Prove that 𝑎 − 𝑏 × 𝑎 + 𝑏 = 2 𝑎 × 𝑏
𝑎 − 𝑏 × 𝑎 + 𝑏
= 𝑎 × 𝑎 − 𝑏 × 𝑎 + 𝑎 × 𝑏 − 𝑏 × 𝑏
= 2 𝑎 × 𝑏
LHS: Vector product of diagonals of parallelogram.
RHS: 2 × Area of parallelogram having 𝑎 and 𝑏 as adjacent sides of the
parallelogram.
𝑎
𝑏
𝑎 − 𝑏
Note

46. Calculate the area of quadrilateral having diagonals 𝑑1 = < 2, 3, −6 >,
𝑑2 = < 3, −4, −1 >.

47. Calculate the area of quadrilateral having diagonals 𝑑1 = < 2, 3, −6 >,
𝑑2 = < 3, −4, −1 >.

48. Vector (or Cross) Product of Two Vectors
If 𝑎 × 𝑏 = 𝑐, 𝑏 × 𝑐 = 𝑎, 𝑐 × 𝑎 = 𝑏 (given), then
𝑎 × 𝑏 = 𝑐 ⇒ 𝑐 ⊥ 𝑎, 𝑐 ⊥ 𝑏
𝑏 × 𝑐 = 𝑎 ⇒ 𝑎 ⊥ 𝑏, 𝑎 ⊥ 𝑐
𝑎 ⊥ 𝑏 ⊥ 𝑐
Now, 𝑎 × 𝑏 = 𝑐 ⇒ 𝑎 × 𝑏 = 𝑐
⇒ 𝑎 𝑏 sin 90° = 𝑐 ⇒ 𝑎 𝑏 = 𝑐 … (𝑖)
Similarly, 𝑏 𝑐 = 𝑎 … (𝑖𝑖)

49. Similarly, 𝑐 = 𝑎 = 1
∴ 𝑎 = 𝑏 = 𝑐 = 1
∴ 𝑎 = 𝑖, 𝑏 = 𝑗, 𝑐 = 𝑘
Three orthogonal unit vectors
𝑎 𝑏
𝑏 𝑐
=
𝑐
𝑎
⇒ 𝑎 = 𝑐
From i , (𝑖𝑖),
∴ 𝑏 = 1 From i

50. Area of Parallelogram
𝑎
𝑏
𝜃
𝑏 sin 𝜃
Area of parallelogram =
1
2
𝑎 𝑏 sin 𝜃 × 2
Area of parallelogram = 𝑎 × 𝑏
Area of parallelogram = 𝑑1 × 𝑑2
𝑑1
𝑑2 𝜃