Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let the d.r's of the normal be $⟨a,b,c⟩$ Equation of the plane is $a(x-0)+b(y+1)+c(z-0)=0$ $\because$ It passes through (0,0,1) $\therefore b+c=0$ Also $\frac{0\cdot a+b-c}{\sqrt{a^2+b^2+c^2\cdot \sqrt{2}}}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ $\Rightarrow b-c=\sqrt{a^2+b^2+c^2}$ And $b+c=0$ $\Rightarrow b=\pm \frac{1}{\sqrt{2}}a$ $\therefore$ The d.r's are $\sqrt{2},1,-1$ or $2,\sqrt{2},-\sqrt{2}$