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Question
Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.
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Solution
The equation of the plane is
6x – 2y + 3z – 7 = 0
∴ its vector equation is
`bar"r".(6hat"i" - 2hat"j" + 3hat"k")` = 7 ...(1)
where `bar"r" = xhat"i" + yhat"j" + zhat"k"`
∴ `bar"n" = 6hat"i" - 2hat"j" + 3hat"k"` is normal to the plane.
`|bar"n"| = sqrt(6^2 + (-2)^2 + 3^2)`
= `sqrt(49)`
= 7
Unit vector along `bar"n"` is
`hat"n" = bar"n"/|bar"n"|= (6hat"i" - 2hat"j" + 3hat"k")/(7)`
Dividing bothsides of (1) by 7, we get
`bar"r".((6hat"i" - 2hat"j" + 3hat"k")/7) = (7)/(7)`
∴ `bar"r".hat"n"`= 1
Comparing with normal form of equation of the plane `hat"r".hat"n" = p` it follows that length of perpendicular from origin is 1 unit.
Alternative Method :
The equation of the plane is 6x – 2y + 3z – 7 = 0
i.e. `(6/(6^2 + (-2)^2 + 3))x - (2/(sqrt(6^2) + (-2)^2 + 3^2))y + ((3)/(sqrt(6^2 + (-2)^2 + 3^2)))z = 7/(sqrt(6^2 + (-2)^2 + 3)`
i.e. `(6)/(7)x -(2)/(7)y + (3)/(7)z = (7)/(7)` = 1
This is the normal form of the equation of plane.
∴ perpendicular distance of the origin frm the plane is p = 1 unit.
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