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The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then

JEE · Math · previous-year question

  1. A.$$a=2,$$ $$b=8$$
  2. B.$$a=4,$$ $$b=6$$
  3. C.$$a=6,$$ $$b=4$$correct
  4. D.$$a=8,$$ $$b=2$$

Answer

C. $$a=6,$$ $$b=4$$

Explanation

Equation of line through $$\left( {5,1,a} \right)$$ and $$\left( {3,b,1} \right)$$ is $${{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda $$ $$\therefore$$ Any point on this line is a $$\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \alpha } \right)\lambda + a} \right]$$ It crosses $$yz$$ plane where $${ - 2\lambda + 5 = 0}$$ $$\lambda = {5 \over 2}$$ $$\therefore$$ $$\left( {0,\left( {b - 1} \right){5 \over 2} + 1,\left( {1 - a} \right){5 \over 2} + a} \right) = \left( {0,{{17} \over 2},{{ - 17} \over 2}} \right)$$ $$ \Rightarrow \left( {b - 1} \right){5 \over 2} + 1 = {{17} \over 2}$$ and $$\left( {1 - a} \right){5 \over 2} + a = - {{13} \over 2}$$ $$ \Rightarrow b = 4$$ and $$a = 6$$

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