If the points $(1, 1, - 3)$ and $(1, 0, - 3)$ lie on opposite sides of the plane $x + y + 3 z + d = 0$ , then

d < 7

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d > 8

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7 <

< 8

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d < 7

d > 8

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Solution

The correct option is B $7 <$ d $< 8$
Putting the value of the points in the equation of plane.
Since, they are on opposite sides, the product will be a negative number.
$(1 + 1 - 9 + d) (1 - 9 + d) < 0$
$(- 7 + d) (- 8 + d) < 0$
$7 < d < 8$

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Similar questions

Identify the points which lie on the line $3 x + 4 y = 7.$

\frac{d^{3} y}{d x^{3}} - 8 \frac{d^{2} y}{d x^{2}} = 0

y (0) = \frac{1}{8}, y^{1} (0) = 0

y^{2} (0) = 1

y^{n} (x) = \frac{d^{n} y}{d x^{n}}]

y = x^{3} + 2 x^{2} + 7 x + 8

\frac{d y}{d x}

$I f 3 x^{2} + 4 x y + 2 y^{2} + x - 8 = 0 t h e n t h e v a l u e o f \frac{d y}{d x} a t (- 1, 3) i s$

3 x + 7 y + 8 = 0

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