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