The direction ratios of normal to the plane through the points (0,-1,0) and (0,0,1) and making an angle…

The direction ratios of normal to the plane through the points (0,-1,0) and (0,0,1) and making an angle $\frac{\pi}{4}$ with the plane $y-z+5=0$ are; 2,-1,1 $2, \sqrt{2}-\sqrt{2}$ $\sqrt{2}, 1,-1$ $2 \sqrt{3}, 1,-1$
  1. option 1 and 2
  2. option 2 and 3
  3. option 3 and 4
  4. all the options

Solution

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

Asked in: JEE Main 2019 (11 Jan Shift 1)

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