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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan 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 $\text{x}+2\text{y}+\lambda\text{z}=5$
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Answer

We know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
The given planes are $2x - 4y + 3z = 5$ and $\text{x}+2\text{y}+\lambda\text{z}=5$
$\Rightarrow a_1 = 2; b_1 = -4; c_1 = 3; a_2 = 1; b_2 = 2; \text{c}_2=\lambda$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow(2)(1)+(-4)(2)+(3)+(\lambda)=0$
$\Rightarrow2-8+3\lambda=0$
$\Rightarrow3\lambda=6$
$\Rightarrow\lambda=2$

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