Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let $\vec{n}$ be inclined at angles $\alpha ,\beta ,\gamma$ to $\mathrm{X},\mathrm{Y},\mathrm{Z}$ axes respectively. $\begin{array}{ll}& \alpha =45^{\circ },\beta =60^{\circ },\gamma =?\\ & {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1\\ \therefore & \frac{1}{2}+\frac{1}{4}+{\cos }^2\gamma =1\\ \therefore & {\cos }^2\gamma =\frac{1}{4}\\ \therefore & \gamma =60^{\circ }\\ & \overrightarrow{\mathrm{n}}=\cos \alpha \hat{\mathrm{i}}+\cos \beta \hat{\mathrm{j}}+\cos \gamma \hat{\mathrm{k}}\\ \therefore & \overrightarrow{\mathrm{n}}=\frac{1}{\sqrt{2}}\hat{\mathrm{i}}+\frac{1}{2}\hat{\mathrm{j}}+\frac{1}{2}\hat{\mathrm{k}}\end{array}$ $\therefore$ Equation of the required plane is $\frac{1}{\sqrt{2}}(x+\sqrt{2})+\frac{1}{2}(y-1)+\frac{1}{2}(z-1)=0$ i.e., $\sqrt{2}x+y+z=0$