How do you write the vector a = 3i + 2j - 6k as a sum of two vectors, one parallel and one perpendicular to d = 2i - 4j + k?

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Answer 1 · 2015-12-15T16:50:58

$lambda d = -8/21 (2, -4, 1)$

$n = 1/21 (79, 10, -118)$

$a = lambda d + n ; and n * d = 0$
$a = ((3), (2), (-6)) ; d = ((2), (-4), (1)) ; n = ((x), (y), (z))$

$n * d = 2x - 4y + z = 0 Rightarrow (2x + z)/4 = y$

$n = ((x), (x/2 + z/4), (z)) = x * ((1), (1/2), (0)) + z * ((0), (1/4), (1))$

$a = lambda d + n Leftrightarrow ((3), (2), (-6)) = lambda * ((2), (-4), (1)) + x * ((1), (1/2), (0)) + z * ((0), (1/4), (1))$

$((3), (8), (-6)) = ((2, 1, 0), (-16, 2, 1), (1, 0, 1)) * ((lambda), (x), (z))$

$L_2 := L_2 - L_3$

$((3), (14), (-6)) = ((2, 1, 0), (-17, 2, 0), (1, 0, 1)) * ((lambda), (x), (z))$

$L_1 := L_2 -2 L_1$

$((8), (14), (-6)) = ((-21, 0, 0), (-17, 2, 0), (1, 0, 1)) * ((lambda), (x), (z))$

$lambda = -8/21$

$14 = 17 * 8/21 + 2x Rightarrow (14 * 21 - 17 * 8)/42 = x = 79/21$

$-6 = -8/21 + z Rightarrow (-6 * 21 + 8)/21 = z = -118/21$

$y = 1/4 (2x + z) = 1/4 * (2*79-118)/21 = 10/21$