SCALAR OR DOT PRODUCT
Direction vector
A vector which is used to specify the direction of another vector can be called a ----- ?
(i + 5j) . (5i - j) = 5 - 5 = 0 Therefore the angle between the two vectors is 90ᵒ since their dot product is zero.
Find the angle between i + 5j and 5i - j.
p . q = |p| |q| cos Ө, -3 = 2 ᵡ 3 cos Ө Ө = arccos (-3/6) = 120ᵒ
Find the angle between the two vectors p = (-√3i + 1j) and q = (0i - 3j)
a . b = (4i + 3j) . (-5i + 2j) = [(4 ᵡ -5) + (3 ᵡ 2)] = -20 + 6 = -14
If vectors; a = 4i + 3j and b = -5i + 2j, find a . b
a . b = (3i + 4j) . (-8i + 6j) = -24 + 24 = 0 Since the dot product of the two vectors is equal to zero, it implies the two vectors a and b are perpendicular.
Show that two vectors a = (3i + 4j) and b = (-8i + 6J) are perpendicular.
The unit vector of the direction vector = 1/5 ᵡ (4i + 3j) Therefore, the vector = magnitude ᵡ unit vector of direction vector = 10 ᵡ 1/5 ᵡ (4i + 3j) = 8i -6j
If a vector has magnitude 10 units and is in the direction of the vector 4i - 3j, find the vector.
Number
The dot product of two vectors give a: ----- ?
a . b = |a| |b| cos Ө 5 = √5 ᵡ √10 cos Ө Ө = arccos (√2/2) = 45ᵒ
Given the vectors a = i + 2j and b = 3i + j, find; the acute angle Ө between a and b.
The projection of the vector a on the vector b, b = (a . b)/|b| b = (5)/√10 = √10/2
Given the vectors a = i + 2j and b = 3i + j, find; the projection of the vector a on the vector b.
a . b = |a| |b| cos Ө
The scalar or dot product of two non-zero vectors a and b is defined as: ----- ?
