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|.cos θ
|a|² + |b|²  = |a|² + |b|²  + 2|a| + |b| ∵−1 ⩽cos θ⩽1
⇒2|a||b|.cos θ ⩽ 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, γare 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 ⁡γare 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