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The distance from $(1, 1, 1)$ to the plane $2x + 2y + z + 3 = 0$ is:
A$8$, the unscaled value of the numerator on chart here
B$1$, taking the normal magnitude alone on the chart
C$2$, dividing the constant by the normal length here
D$8/3$, by $|2 + 2 + 1 + 3|/\sqrt{4 + 4 + 1} = 8/3$
Answer & Solution
Correct answer: D. $8/3$, by $|2 + 2 + 1 + 3|/\sqrt{4 + 4 + 1} = 8/3$
$d = |2(1) + 2(1) + 1(1) + 3|/\sqrt{4 + 4 + 1} = 8/3$.
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