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प्रश्न
Find the distance of the point `4hat"i" - 3hat"j" + hat"k"` from the plane `bar"r".(2hat"i" + 3hat"j" - 6hat"k")` = 21.
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उत्तर
Given:
Point P = `4hat"i" - 3hat"j" + hat"k"`
Plane: `bar"r".(2hat"i" + 3hat"j" - 6hat"k") = 21`
2x + 3y − 6z = 21
`D = |ax_0 + by_0 + cz_0 + d|/(sqrta^2 + b^2 + c^2)`
2x + 3y − 6z = 21 ⇒ 2x + 3y − 6z − 21 = 0
`4hati - 3hatj + hatk => (x_0, y_0, z_0) = (4, −3, 1)`
`D = |2(4) + 3(-3) + (-6)(1) - 21|/sqrt(2^2 + 3^2 + (-6)^2)`
`D = |8-9-6-21|/sqrt(4+9+36) = |-28|/sqrt49 = 28/7`
= 4 units
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