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The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are :

JEE · Math · previous-year question

  1. A.2, $$-$$1, 1
  2. B.$$2\sqrt 3 ,1, - 1$$
  3. C.$$\sqrt 2 ,1, - 1$$correct
  4. D.$$\sqrt 2 , - \sqrt 2 $$

Answer

C. $$\sqrt 2 ,1, - 1$$

Explanation

Let the equation of plane be a(x $$-$$ 0) + b(y + 1) + c(z $$-$$ 0) = 0 It passes through (0, 0, 1) then b + c = 0 . . . . (1) Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$$ $$ \Rightarrow $$ a2 $$=$$ $$-$$ 2bc and b $$=$$ $$-$$ c we get a2 $$=$$ 2c2 $$ \Rightarrow $$ a $$=$$ $$ \pm $$ $$\sqrt 2 $$ c $$ \Rightarrow $$ direction ratio (a, b, c) = ($$\sqrt 2 $$, $$-$$1, 1) or ($$\sqrt 2 $$, 1, $$-$$ 1)

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