If the distance of points 2 i ^ + 3 j ^ + λ k ^ from the plane r → ⋅ ( 3 i ^ + 2 j ^ + 6 k ^ ) = 13 is 5 units then λ =

Equation of plane r ¯ ⋅ 3 i ^ + 2 j ^ + 6 k ^ = 13 i . e . 3 x + 2 y + 6z - 13 = 0 Given point ( 2 , 3 , λ ) Distance of plane from the point = a x 1 + b y 1 + c z 1 + d a 2 + b 2 + c 2 ⇒ 5 = 3 2 + 2 3 + 6 λ - 13 9 + 4 + 36 ⇒ 5 = 6 λ - 1 7 ⇒ 6 λ - 1 = ± 35 ⇒ 6 λ = 36 , 6 λ = - 34 λ = 6 , λ = - 17 3

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Mathematics3D Coordinate GeometryPlane (Equations)Medium2 minPYQ_2017

MathematicsMediumsingle choice

If the distance of points $2 \hat{i} + 3 \hat{j} + λ \hat{k}$ from the plane $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$ is $5$ units then $λ =$

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Equation of plane $\bar{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$
$i . e . 3 x + 2 y + 6z - 13 = 0$
Given point $(2, 3, λ)$
Distance of plane from the point $= |\frac{a x_{1} + b y_{1} + c z_{1} + d}{\sqrt{a^{2} + b^{2} + c^{2}}}|$
$\Rightarrow$ $5 = |\frac{3 (2) + 2 (3) + 6 λ - 13}{\sqrt{9 + 4 + 36}}|$
$\Rightarrow$ $5 = |\frac{6 λ - 1}{7}|$
$\Rightarrow 6 λ - 1 = \pm 35$
$\Rightarrow 6 λ = 36, 6 λ = - 34$
$λ = 6, λ = - \frac{17}{3}$

Stream:MHT_CETSubject:MathematicsTopic:3D Coordinate GeometrySubtopic:Plane (Equations)

⏱ 2mℹ️ Source: PYQ_2017

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