हिंदी

Find the distance of the point (1, 1 –1) from the plane 3x +4y – 12z + 20 = 0.

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प्रश्न

Find the distance of the point (1, 1 –1) from the plane 3x +4y – 12z + 20 = 0.

योग
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उत्तर

The distance of the point (x1, y1, z1) from the plane ax + by + cz + d = 0 is `|(ax_1 + by_1 + cz_1 + d)/sqrt(a^2 + b^2 + c^2)|`

∴ the distance of the point (1, 1, – 1) from the plane 3x + 4y – 12z + 20 = 0 is `|(3(1) + 4(1 - 12(-1) + 20))/sqrt(3^2 + 4^2 + (-12)^2)|`

= `|(3 + 4 + 12 + 20)/sqrt(9 + 16 + 144)|`

= `(39)/sqrt(169)`

= `(39)/(13)`

= 3units.

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अध्याय 6: Line and Plane - Exercise 6.4 [पृष्ठ २२०]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 6 Line and Plane
Exercise 6.4 | Q 5 | पृष्ठ २२०

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