Vector v4

Find the vector v4 perpendicular to the vectors
v1 = (1, 1, 1, -1), v2 = (1, 1, -1, 1) and v3 = (0, 0, 1, 1)

Correct answer:

v₄ = (1,-1,0,0)

v_{4} \cdot v_{1} = 0 v_{4} \cdot v_{2} = 0 v_{4} \cdot v_{3} = 0 1 x + 1 y + 1 z - 1 w = 0 1 x + 1 y - 1 z + 1 w = 0 0 x + 0 y + 1 z + 1 w = 0 x = 1 1 \cdot x + 1 \cdot y + 1 \cdot z - 1 \cdot w = 0 1 \cdot x + 1 \cdot y - 1 \cdot z + 1 \cdot w = 0 0 \cdot x + 0 \cdot y + 1 \cdot z + 1 \cdot w = 0 x = 1 w - x - y - z = 0 w + x + y - z = 0 w + z = 0 x = 1 R o w 2 - R o w 1 \to R o w 2 w - x - y - z = 0 2 x + 2 y = 0 w + z = 0 x = 1 R o w 3 - R o w 1 \to R o w 3 w - x - y - z = 0 2 x + 2 y = 0 x + y + 2 z = 0 x = 1 R o w 3 - \frac{1}{2} \cdot R o w 2 \to R o w 3 w - x - y - z = 0 2 x + 2 y = 0 2 z = 0 x = 1 R o w 4 - \frac{1}{2} \cdot R o w 2 \to R o w 4 w - x - y - z = 0 2 x + 2 y = 0 2 z = 0 - y = 1 P i v o t : R o w 3 \leftrightarrow R o w 4 w - x - y - z = 0 2 x + 2 y = 0 - y = 1 2 z = 0 z = \frac{0}{2} = 0 y = \frac{1}{- 1} = - 1 x = \frac{0 - 2 y}{2} = \frac{0 - 2 \cdot ( - 1 )}{2} = 1 w = 0 + x + y + z = 0 + 1 - 1 = 0 w = 0 x = 1 y = - 1 z = 0 v_{4} = (1, - 1, 0, 0)

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