Vector and Scalar Projections

Scalar Projection

  • The scalar projection of vector a in the direction of vector b represents the component of a that lies along b.
  • It can be visualized as the "shadow" of a onto b.
  • Denoted as d, the scalar projection can be calculated using trigonometry: d=∣a∣cos⁡θ=∣a∣a⋅b∣a∣∣b∣=a⋅b∣b∣d = |a| \cos \theta = |a| \frac{a \cdot b}{|a||b|} = \frac{a \cdot b}{|b|}
    • dd is the scalar projection.
    • θ\theta is the angle between vectors a and b.
    • ∣a∣|a| and ∣b∣|b| are the magnitudes of vectors a and b, respectively.

Worked Example 1

  • Problem: Find the scalar projection of a=i+3j−6ka = i + 3j - 6k in the direction of b=4i+2j+4kb = 4i + 2j + 4k.
  • Solution:
    d=a⋅b∣b∣=(1,3,−6)⋅(4,2,4)6=4+6−246=−146=−73d = \frac{a \cdot b}{|b|} = \frac{(1, 3, -6) \cdot (4, 2, 4)}{6} = \frac{4 + 6 - 24}{6} = \frac{-14}{6} = -\frac{7}{3}
  • The negative value of d indicates that the angle θ\theta between the vectors is greater than 90 degrees.

Vector Projection

  • The vector projection of a onto b is a vector, v, that lies along b and has a magnitude equal to d (the scalar projection).
  • Given the magnitude d and the direction (defined by the unit vector b^\hat{b}), the vector projection v can be expressed as:
    v=db^=(a⋅b^)b^v = d \hat{b} = (a \cdot \hat{b}) \hat{b}

Worked Example 2

  • Problem: Determine the scalar and vector projections of a=i+3j+6ka = i + 3j + 6k in the direction of b=i+2j+2kb = i + 2j + 2k.
  • Scalar Projection:
    d=a⋅b∣b∣=(1,3,6)⋅(1,2,2)3=1+6+123=193d = \frac{a \cdot b}{|b|} = \frac{(1, 3, 6) \cdot (1, 2, 2)}{3} = \frac{1 + 6 + 12}{3} = \frac{19}{3}
  • Vector Projection:
    b^=b∣b∣=(1,2,2)3=(13,23,23)\hat{b} = \frac{b}{|b|} = \frac{(1, 2, 2)}{3} = (\frac{1}{3}, \frac{2}{3}, \frac{2}{3})
    v=db^=193(13,23,23)=(199,389,389)v = d \hat{b} = \frac{19}{3} (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) = (\frac{19}{9}, \frac{38}{9}, \frac{38}{9})

Resolution into Perpendicular Components

  • A vector a can be resolved into two components relative to another vector b: one parallel to b (the vector projection v) and one perpendicular to b (vector w).
  • Since v+w=av + w = a, the perpendicular component w can be found by:
    w=a−vw = a - v

Worked Example 3

  • Problem: Resolve the vector a=i+3j+6ka = i + 3j + 6k into two components, one parallel to b=i+2j+2kb = i + 2j + 2k and the other perpendicular to b.
  • Solution:
    • We already found v=(199,389,389)v = (\frac{19}{9}, \frac{38}{9}, \frac{38}{9}) in Worked Example 2.
    • Now, we find w:
      w=a−v=(1,3,6)−(199,389,389)=(99,279,549)−(199,389,389)=(−109,−119,169)w = a - v = (1, 3, 6) - (\frac{19}{9}, \frac{38}{9}, \frac{38}{9}) = (\frac{9}{9}, \frac{27}{9}, \frac{54}{9}) - (\frac{19}{9}, \frac{38}{9}, \frac{38}{9}) = (\frac{-10}{9}, \frac{-11}{9}, \frac{16}{9})
  • Verification:
    • To check that v and w are perpendicular, verify that their dot product is zero. Since v is a multiple of b, it is sufficient to check the dot product of b and w:
      (1,2,2)⋅(−109,−119,169)=(1,2,2)⋅(−10,−11,16)=−10−22+32=0(1, 2, 2) \cdot (\frac{-10}{9}, \frac{-11}{9}, \frac{16}{9}) = (1, 2, 2) \cdot (-10, -11, 16) = -10 - 22 + 32 = 0
  • Final Resolution:
    a=v+w=(199,389,389)+(−109,−119,169)a = v + w = (\frac{19}{9}, \frac{38}{9}, \frac{38}{9}) + (\frac{-10}{9}, \frac{-11}{9}, \frac{16}{9})