Question

If the normal of the plane makes angles $\frac{\pi }{4},\frac{\pi }{4}\text{and}\frac{\pi }{2}$ with positive $X$-axis and $Y$-axis and $Z$-axis respectively and the length of the perpendicular line segment from origin to the plane is $\sqrt{2}$, then the equation of the plane is:

• A. $x+y+z=1$
• B. $x+y=2$
• C. $x=\sqrt{2}$
• D. $x+y+z=\sqrt{2}$

Answer 1

The correct option is B $x+y=2$

Normal of the plane makes an angeles $\frac{\pi }{4},\frac{\pi }{4}\text{and}\frac{\pi }{2}$ with positive $X$-axis and $Y$-axis and $Z$-axis respectively,
Direction cosine of normal to the plane is $(\cos \alpha ,\cos \beta ,\cos \gamma )$
Direction cosine of normal to the plane is
$(\cos \frac{\pi }{4},\cos \frac{\pi }{4},\cos \frac{\pi }{2})=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$
Equation of the plane is given by
$x\cos \alpha +y\cos \beta +z\cos \gamma =P$
$(P=\perp \text{distance from origin to the plane})$.
$\therefore \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\sqrt{2}$
$\Rightarrow x+y=2$