Question

what is the distance of point P(3,4,-5) from? (I)plane z = 6 (ii) plane x = 7​

Answer 1

Step-by-step explanation:

$distance \: of \: pt \: (\alpha . \beta . \gamma ) \\ from \: ax + by + cz + d = 0 \\ is \: = |\frac{a \alpha + b \beta + c \gamma + d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } } | \\ therefore \: distance \: of \: (3.4. - 5) \\ \: from \: z - 6 = 0 \: is \: | - 5 - 6| = 11 \\ distance \: from \: x - 7 = 0 \\ \: is \: |3 - 7| = 4$