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The plane $2x - y + 3z + 5 = 0$ has normal vector:
A$(5, 0, 0)$, treating the constant as the normal on chart
B$(2, -1, 3)$, the coefficients of $x, y, z$ in plane formula
C$(0, 0, 5)$, taking the constant as the $z$-component here
D$(2, 1, 3)$, taking absolute values without keeping sign
Answer & Solution
Correct answer: B. $(2, -1, 3)$, the coefficients of $x, y, z$ in plane formula
For $ax + by + cz + d = 0$, the normal is $(a, b, c) = (2, -1, 3)$.
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