Test: Introduction To Vector Algebra - JEE MCQ

Two vectors are equal if their corresponding components are equal.
Hence, 2x = 3
or x = 3/2
and 2x = y
or y = 2 * 3/2
y = 3

Two or more vectors having the same initial point are called coinitial vectors.

Both of these vectors are pointing the same direction but they have different magnitude. Hence, they are collinear vectors.

|a| = (x² + 2² + (2x)²)^1/2
7 = (x² + 2² + (2x)²)^1/2
⇒ 49 = x2 + 2² + 4x²
⇒ 49 = 4 + 5x²
⇒ 5x² = 45
⇒ x² = 9
x = ±3

Given:

If we divide both sides with 2 we get,

⇒ Mid pt. of AC = Mid. pt. of BD
∴ ABCD is a parallelogram.

a = 2i + 3j - 6k
|a| = √4+9+36 = √49 = 7
b = 6i - 2j + 3k
|b| = √36+4+9 = √49 = 7
|a| = |b|
Hence, option A is correct.

In the Additive Identity of vectors, the additive identity is zero vector 0.
For any vector V additive identity is defined as,
0 + V = V and V + 0 = V

Direction Cosines for angle α, β, γ are:
cos α, cos β, cos γ

Position vector of A = OA = a
Position vector of B = OB = b
Now, AC = OC - OA
Also, AB = OB - OA
Given, AC = 3AB
⇒ OC - OA = 3(OB - OA)
⇒ OC = 3b - 2a

A vector whose magnitude (i.e., length) is equal to 1 is called a unit vector. There are exactly two unit vectors in any given direction and one is the negative of the other.

Given, a = i + 2j
b = -2i + j
c = 4i +3j
Also, c = xa +yb
Now putting the values in above equation,
4i + 3j  = x(i + 2j) + y(-2i +j)
⇒ xi + 2xj - 2yi + yj
⇒ (x-2y)i + (2x+y)j
We get,
x - 2y = 4
2x + y = 3
After solving,            
x = 2
y = -1

Zero vector is the unit vector having zero length, hence the direction is undefined.

Given, Point A (2,-3,7)

Point B (1,3,-4)

Let vector in the direction of AB be C.

∴ C = B - A

⇒ (1,3,-4) - (2,-3,7)

⇒ ( 1-2 , 3+3 , -4-7 )

⇒ (-1,6,-11)

⇒ -1i + 6j -11k
Magnitude of vector C
|C| = √(-1)² + 6² + (-11)²
⇒ √1+36+121
⇒ √158

Unit vector = (Vector)/(Magnitude of vector)
Unit vector C = (C vector)/(Magnitude of C vector)  = (-1i + 6j -11k)/√158

Since a is the magnitude of the vector, it is always positive and it can be 0 in case of zero vectors.        
So, a ≥ 0

Given vector = i + 2j - 3k
Magnitude = √1² + 2² + (-3)² = √14
Unit vector in direction of resultant = (i + 2j - 3k) / √14
Vector of magnitude 14​ unit in direction of resultant,
⇒ 14[ (i + 2j - 3k) / √14 ]
⇒ √14(i + 2j - 3k)

|a + b|² = |a|² + |b|² + 2|a||b|.cosθ
|a|² + |b|² = |a|² + |b|² + 2|a| + |b|  ∵ −1 ⩽ cosθ ⩽ 1
⇒ 2|a||b|.cosθ ⩽ 2|a||b|
So, |a + b|² ⩽ (|a| + |b|)²
⇒ |a + b| ≤ |a| + |b|
This is also known as Triangle Inequality of vectors.

Consider   is the position vector of a point M(x,y,z) and α, β, γ are the angles, made by the vector   with the positive directions of x, y and z respectively. The cosines of the angles, cos⁡α, cos⁡β, cos⁡γ are the direction cosines of the vector   denoted by l, m, n, then
cos²⁡α + cos²⁡β+ cos²⁡γ =1  i.e.l² + m² + n² = 1.

A vector whose initial and terminal points coincide has no particular direction and 0 magnitude. Therefore, it is called zero vector.

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point.

Given a = (-1,1) and b = (m,-2)
Given that above two vectors are collinear, so they are parallel
⇒ -1/m = 1/-2
⇒ m = 2