If the points 1-1- - and -3-0-1 are equidistant from the plane- 3 x -4 y -12 z -13-0- then - satisfies the equation-A- 3 x2-10 x-21-0B- 3 x2-10 x-13-0- 3 x2-10 x-7-0D- 3 x 2-10 x -7-0

The correct option is C 3x^2 10x+7=0 nbsp; nbsp; nbsp;|3·1+4 · 1 12 ·λ +13/√(3^2+4^2+( 12)^2)|=|3 ( 3) 12 · 1 +13/√(3^2+4^2+( 12)^2)| ⇒| 20 12 λ|=8 ⇒ 2 ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Plane

If the points...

Question

If the points $(1, 1, λ)$ and $(- 3, 0, 1)$ are equidistant from the plane, $3 x + 4 y - 12 z + 13 = 0,$ then $λ$ satisfies the equation:

3 x^{2} - 10 x + 21 = 0

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3 x^{2} + 10 x - 13 = 0

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3 x^{2} - 10 x + 7 = 0

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3 x^{2} + 10 x + 7 = 0

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Solution

The correct option is C $3 x^{2} - 10 x + 7 = 0$
$∣ ∣ ∣ ∣ \frac{3 \cdot 1 + 4 \cdot 1 - 12 \cdot λ + 13}{\sqrt{3^{2} + 4^{2} + (- 12)^{2}}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \frac{3 (- 3) - 12 \cdot 1 + 13}{\sqrt{3^{2} + 4^{2} + (- 12)^{2}}} ∣ ∣ ∣ ∣$
$\Rightarrow | 20 - 12 λ | = 8$
$\Rightarrow 20 - 12 λ = \pm 8$
$\Rightarrow λ = 1, \frac{7}{3}$
$\Rightarrow λ$ satisfies the equation $x^{2} - (1 + \frac{7}{3}) x + \frac{7}{3} = 0$
$i.e., 3 x^{2} - 10 x + 7 = 0$

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If the points (1,1,λ) and (−3,0,1) are equidistant from the plane, 3x+4y−12z+13=0, then λ satisfies the equation:

The correct option is C 3x2−10x+7=0 ∣∣ ∣∣3⋅1+4⋅1−12⋅λ+13√32+42+(−12)2∣∣ ∣∣=∣∣ ∣∣3(−3)−12⋅1+13√32+42+(−12)2∣∣ ∣∣ ⇒|20−12λ|=8 ⇒20−12λ=±8 ⇒λ=1,73 ⇒λ satisfies the equation x2−(1+73)x+73=0 i.e., 3x2−10x+7=0

If the points $(1, 1, λ)$ and $(- 3, 0, 1)$ are equidistant from the plane, $3 x + 4 y - 12 z + 13 = 0,$ then $λ$ satisfies the equation: