Question

The planes $3x-6y-2z=7\mathrm{and}2x+y-\lambda z=5$ are perpendicular. Then the value of $\lambda$ is __________.

Answer 1

Given, planes, $3x-6y-2z=7\mathrm{and}2x+y-\lambda z=5$ are perpendicular

$$\begin{gathered} \mathrm{Since}\mathrm{two}\mathrm{planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\mathrm{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\mathrm{are}\mathrm{perpendicular} \\ \mathrm{If}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0 \\ \mathrm{i}.\mathrm{e}.3\left(2\right)+\left(-6\right)\left(1\right)+\left(-2\right)\left(-\lambda \right)=0 \\ \mathrm{i}.\mathrm{e}.6-6+2\lambda =0 \\ \mathrm{i}.\mathrm{e}.2\lambda =0 \\ \mathrm{i}.\mathrm{e}.\lambda =0 \end{gathered}$$