%%

A perpendicular is drawn from a point on the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$$ to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :

JEE · Math · previous-year question

  1. A.(4, 0, – 1)
  2. B.(2, 0, 1)correct
  3. C.(1, 0, 2)
  4. D.(– 1, 0, 4)

Answer

B. (2, 0, 1)

Explanation

$${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1} = \lambda $$ Let a point P on the line is (2$$\lambda $$ + 1, – $$\lambda $$ –1, + $$\lambda $$) Foot of $${ \bot ^r}Q$$ is given by $${{x - 2\lambda - 1} \over 1} = {{y + \lambda + 1} \over 1} = {{z - \lambda } \over 1} = - {{\left( {2\lambda - 3} \right)} \over 3}$$ $$ \therefore $$ Q lies on x + y + z = 3 & x – y + z = 3 $$ \Rightarrow $$ x + z = 3 & y = 0 $$ \therefore $$ $$y = 0 \Rightarrow \lambda + 1 = {{ - 2\lambda + 3} \over 3} \Rightarrow \lambda = 0$$ $$ \therefore $$ Q is (2, 0, 1)

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions