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The plane — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThe plane3 Marks
Question
Determine the value of $\lambda$ for which the following planes are perpendicular to other. $3\text{x}-6\text{y}-2\text{z}=7$ and $2\text{x}+\text{y}-\lambda\text{z}=5$
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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 $3x - 6y - 2z = 7$ and $2\text{x}+\text{y}-\lambda\text{z}=5$
$\Rightarrow a_1 = 3; b_1 = -6; c_1 = -2; a_2 = 2; b_2 = 1;$
$\text{c}_2=-\lambda$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow(3)(2)+(-6)(1)+(-2)(-\lambda)=0$
$\Rightarrow6-6+2\lambda=0$
$\Rightarrow2\lambda=0$
$\Rightarrow\lambda=0$

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