
Vector-Algebra
VECTOR ALGEBRA
TopConcepts
The point X from where the vector starts is called the initial point and the point Y where it ends is called the terminal point.
In general, l2 + m2 + n2 = 1, but a2 + b2 + c2 ≠ 1.
Thus, if →a is a vector with magnitude a, then ∣∣a−1––––∣∣=1a.
Thus, (i) Commutative: For any two vectors →a and →b , we have
→a + →b = →b + →a
(ii) Associativity: For any three vectors, →a, →b, and →c , we have
k > 0 ⇒ k →ahas the same direction as →a.
k < 0 ⇒ k →a has the opposite direction as →a.
Magnitude of k →a is |k| times the magnitude of vector k→a.
Vector −−→OP = →r = xˆi+ yˆj + zˆk is called the component form of vector r.
Here, x, y and z are called the scalar components of →r in the directions of and ˆi,ˆj and ˆk, and
xˆi, yˆj and zˆk are called the vector components of the vector r along the respective axes.

l→a + m→b + n→c = →0, where I, m and n are scalars not all zero simultaneously.





Top Formulae



Then the components of →balong and perpendicular to →a are




Let →a and b be two non-zero, non-parallel vectors.
Then the cross-product →a × →b is a vector with magnitude equal to the area of the parallelogram having →a and →b as its adjacent sides and direction ˆn is perpendicular to the plane of →a and →b.
Important Questions
Multiple Choice questions-
1. In ΔABC, which of the following is not true?
(a) −−→AB + −−→BC + −−→CA = →0
(b) −−→AB + −−→BC – −−→AC = →0
(c) −−→AB + −−→BC – −−→CA = →0
(d) −−→AB – −−→CB + −−→CA = →0
2. If →a and →b are two collinear vectors, then which of the following are incorrect:
(a) →b = λ→a tor some scalar λ.
(b) →a = ±→b
(c) the respective components of →a and →b are proportional
(d) both the vectors →a and →b have the same direction, but different magnitudes.
3. If a is a non-zero vector of magnitude ‘a’ and λa non-zero scalar, then λ→a is unit vector if:
(a) λ = 1
(b) λ = -1
(c) a = |λ|
(d) a = 1|λ|
4. Let λ be any non-zero scalar. Then for what possible values of x, y and z given below, the vectors 2ˆi – 3ˆj + 4ˆk and xˆi – yˆj + zˆk are perpendicular:
(a) x = 2λ. y = λ, z = λ
(b) x = λ, y = 2λ, z = -λ
(c) x = -λ, y = 2λ, z = λ
(d) x = -λ, y = -2λ, z = λ.
5. Let the vectors →a and →b be such that |→a| = 3 and |→b| = √23, then →a×→b is a unit vector if the angle between →a and →b is:
(a) π6
(b) π4
(c) π3
(d) π2
6. Area of a rectangle having vertices
A (-ˆi + 12ˆj + 4ˆk),
B (ˆi + 12ˆj + 4ˆk),
C (ˆi – 12ˆj + 4ˆk),
D (-ˆi – 12ˆj + 4ˆk) is
(a) 12 square unit
(b) 1 square unit
(c) 2 square units
(d) 4 square units.
7. If θ is the angle between two vectors →a, →b, then →a.→b≥ 0 only when
(a) 0 < θ <π2
(b) 0 ≤ θ ≤ π2
(c) 0 < θ < π
(d) 0 ≤ θ ≤ π
8. Let →a and →b be two unit vectors and 6 is the angle between them. Then →a + →b is a unit vector if:
(a) θ = π4
(b) θ = π3
(c) θ = π2
(d) θ = 2π3
9. If {ˆi, ˆj, ˆk} are the usual three perpendicular unit vectors, then the value of:
ˆi.(ˆj × ˆk) + ˆj.(ˆi × ˆk) + ˆk.(ˆi × ˆj) is
(a) 0
(b) -1
(c) 1
(d) 3
10. If θ is the angle between two vectors →a and →b, then |→a.→b| = |→a×→b| when θ is equal to:
(a) 0
(b) π4
(c) π2
(d) π
Very Short Questions:
(i) 40°
(ii) 50 watt
(iii) 10gm/cm3
(iv) 20 m/sec towards north
(v) 5 seconds. (N.C.E.R.T.)
(i) Collinear
(ii) Equal
(iii) Co-initial. (N.C.E.R.T.)
Short Questions:
(→a− →b) are perpendicular to each other. (C.B.S.E. Outside Delhi 2019)
is √3. (C.B.S.E. 2019)
(C.B.S.E. Sample Paper 2019-20)
Long Questions:
Case Study Questions:
1. A barge is pulled into harbour by two tug boats as shown in the figure.
Based on the above information, answer the following questions.
2. Three slogans on chart papers are to be placed on a school bulletin board at the points A, Band C displaying A (Hub of Learning), B (Creating a better world for tomorrow) and C (Education comes first). The coordinates of these points are (1, 4, 2), (3, -3, -2) and (-2, 2, 6) respectively.
Based on the above information, answer the following questions.
Answer Key-
Multiple Choice questions-
Very Short Answer:
(i) Angle-scalar
(ii) Power-scalar
(iii) Density-scalar
(iv) Velocity-vector
(v) Time-scalar.
(i) →a, →c and →a are collinear vectors.
(ii) →a and →c are equal vectors.
(iii) →b, →c and →d are co-initial vectors.
Sum of the vectors = ˆa + ˆb + ˆc
Since the vector is directed from P to Q,
∴ P is the initial point and Q is the terminal point.
∴ Reqd. vector = −−→PQ
Here
Comparing, A: = 3,2 = -y i.e. y = -2, ̴ z= 1 i.e. z = -1.
Hence, x + y + z = 3 – 2 – 1 = 0.
The given vectors 3ˆi + 2ˆj + 9ˆk and ˆi − 2pˆj + 3ˆk are parallel
Short Answer:
Position vector of
Hence, the unit vector perpendicular to both →a and →b
We have:
⇒ (2 –λ) (3) + (2 + 2λ) (1) + (3 + λ)(0) = 0
⇒ 6 – 3λ + 2 + 2λ = 0
⇒ –λ, + 8 = 0.
Hence, λ, = 8.
Now,
= (4) (- 2) + (1) (3) + (-1) (- 5)
= – 8 + 3 + 5 = 0.
By the question,
⇒ΔABC is equilateral, each of its angles being 60°
⇒∠DAB = 2 x 60° = 120° and ∠ADB = 30°.
By Sine Formula,
Long Answer:
⇒ 3x + y – z = 0 …(1)
and x – 4y + 5z = 0 …(2)
⇒ 4x + 5y-z = 21 …(3)
Multiplying (1) by 5,
1 5x + 5y – 5z = 0 …(4)
Adding (2) and (4),
16x + y = 0 …(5)
Subtracting (1) from (3),
x + 4y = 21 …(6)
From (5),
y = -16x …(7)
Putting in (6),
x – 64x = 21
-63x: = 21
z = 13/3
(1) (a) + (-2) (b) + 1 (c) = 0
⇒ a – 2b + c = 0
⇒ (1) (-2c – b) – (1) (c – a) + (1) (b + 2a) = 0
⇒ -2c-b-c + a + b + 2a = 0
⇒ 3a – 3c = 0
⇒ a – c – 0
Solving (1) and (2),
Note: If ‘θ’ is the angle between AB and CD,
where ‘θ’ is the angle between a and b
⇒ (3)2 + (5)2 + 2 (3) (5) cos θ = (7)2
⇒9 + 25 + 30 cos θ = 49
⇒ 30 cos θ = 49 – 34 ⇒ cos θ = 12
⇒θ = 60°.
Case Study Answers:
1. Answer :
2. Answer :