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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry3 Marks
Question
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$2\text{x}-4\text{y}+3\text{z}=5$ and $\text{x}+2\text{y}+\lambda\text{z}=5$
✓

Answer

We know that the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are perperndicular to each other only if
a1a2 + b1b2 + c1c2 = 0
The given planes are 2x - 4y + 3z = 5 and $\text{x}+2\text{y}+\lambda\text{z}=5$
⇒ a1 = 2; b1 = -4; c1 = 3; a2 = 1; b2 = 2; $\text{c}_2=\lambda$
It is given that the given planes are perpendicular.
⇒ a1a2 + b1b2 + c1c2 = 0
$\Rightarrow(2)(1)+(-4)(2)+(3)+(\lambda)=0$
$\Rightarrow2-8+3\lambda=0$
$\Rightarrow3\lambda=6$
$\Rightarrow\lambda=2$

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