Exercise 1: Decomposing a=(6,2) Onto b=(1,3) — Possible Solution
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COMPUTING THE SCALAR PROJECTION FACTOR
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a = (6,2), b = (1,3).

    a.b = 6*1 + 2*3 = 6 + 6 = 12
    b.b = 1*1 + 3*3 = 1 + 9 = 10
    scalar = a.b / b.b = 12/10 = 1.2

COMPUTING THE PARALLEL COMPONENT (PROJECTION)
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    proj_b(a) = scalar * b = 1.2 * (1,3) = (1.2, 3.6)

COMPUTING THE PERPENDICULAR COMPONENT
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    perp = a - proj_b(a) = (6,2) - (1.2,3.6) = (4.8, -1.6)

VERIFICATION 1: THE TWO PARTS SUM BACK TO a
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    proj + perp = (1.2 + 4.8, 3.6 + (-1.6)) = (6.0, 2.0)

This matches the original vector a = (6,2) exactly (computed in
floating point, the components come out as 1.2 and 3.5999999999999996
and 4.8 and -1.5999999999999996, which sum back to exactly 6.0 and
2.0 - a tiny rounding artifact in the intermediate values that
cancels out in the sum).

VERIFICATION 2: THE PERPENDICULAR COMPONENT IS ORTHOGONAL TO b
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    perp . b = 4.8*1 + (-1.6)*3 = 4.8 - 4.8 = 0

Computed precisely in floating point, this comes out to
8.88*10^-16 - not exactly zero due to ordinary floating-point
rounding (per Numerical Methods & Floating-Point Computation's own
material), but close enough to zero to confirm genuine orthogonality
rather than a calculation error.

WHY THIS WORKS AS AN ANSWER
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The answer follows this chapter's own projection formula step by
step, arrives at the two component vectors, and performs both of the
chapter's own recommended verification checks (the sum recovering the
original vector, and the dot product confirming orthogonality) rather
than stopping at the raw formula result without confirming it.