For each of the following problems, sketch the vectors involved on a coordinate plane, then decompose vector into two vectors →v₁ and →v₂, where →v₁ is parallel to →w, and →v₂ is orthogonal to →w.

Question

1. →v=- →i+6→j
2. →w=3→i-2→j

Answer 1

Given:

Two vectors:

v = - i + 6 j

w = 3 i - 2 j

Where bold letters are vectors, i and j are orthogonal unit vectors, say in x and y direction.

Required:

Projection of v on w?

Analysis:

(1) Projection of v on w is obtained by the dot product v•w / magnitude |w|

v • w / |w|= ( - i + 6 j ) •( 3 i - 2 j ) = (-1)(3) +(6)(-2) = -15 / |w| ------------(1)

Thus, the magnitude of the projection of v on w = -15/ |w|

(2) The unit vector along w is obtained by dividing w by its magnitude |w|.

Magnitude of w = sqrt ( w • w )

........................= sqrt ( (3 i - 2 j )(3 i - 2 j ))

........................= sqrt ( (9+4)

................|w|...= sqrt ( (13) ----------------------------------(2)

The unit vector along w is = w / |w|

..........................................= (3 i - 2 j ) / sqrt (13)-------(3)

(3) The projection of v on w = ( I v • w | / |w • w| ) w

From equation (1), |v • w| = -15

From equation (2), |w| = sqrt(13) or |w • wI = 13

Therefore,

( I v • w | / |w • w| ) w = ( -15 /13 ) w

------------------------------ = ( -15 /13 ) (3 i - 2 j ) ---------------------(4)

Equation (14) gives the required vector v₁ that is parallel to w.

v₁ = ( -15 /13 ) (3 i - 2 j )

(4) The perpendicular of v on v₂ = v - v₁

We already have v as (- i + 6 j ) from statement of problem.

We also derived v₁ = ( -15 /13 ) (3 i - 2 j ) , equation (4).

Therefore,

v₂ = v - v₁

.... = (- i + 6 j ) - ( -15 /13 ) (3 i - 2 j )

.....= (- 1 + 3(15/13) ) i + ( 6 - 2(15/13)) j

.....= (32/13) i + ( 48/13) j --------------------------(5)

Equation (15) gives the perpendicular of v on w.

SUMMARY

Two vectors:

v = - i + 6 j

w = 3 i - 2 j

Projections:

v₁ = ( -15 /13 ) (3 i - 2 j )

v₂ = (32/13) i + ( 48/13) j