"A line 'l' passing through origin is perpendicular to the lines$${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$If the co-ordinates of the point in the first octant on 'l2‘ at a distance of $\sqrt {17}$ from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to ___________."

Question

Accepted Answer

"$${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$

${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow$ D.R. of ${l_1} = 1,2,2$

$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$

${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow$ D.R. of ${l_2} = 2,2,1$

D.R. of l is $\bot$ to l₁ & k₂

$\therefore$ D.R. of $l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$

$\therefore$ Equation of $l:{x \over 2} = {y \over { - 3}} = {z \over 2}$

Solving l & l₁

$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$

$\Rightarrow 2\lambda = \mu + 3$

$- 3\lambda = 2\mu - 1$

$2\lambda = 2\mu + 4$

$\Rightarrow \mu + 3 = 2\mu + 4$

$\mu = - 1$

$\lambda = 1$

$P(2, - 3,2)$ {intersection point}

Let, $Q(2v + 3,2v + 3,v + 2)$ be point on l₂

Now, $$PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $$

$\Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$

$\Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$

$\Rightarrow 9{v^2} + 28v + 20 = 0$

$\Rightarrow 9{v^2} + 18v + 10v + 20 = 0$

$\Rightarrow (9v + 10)(v + 2) = 0$

$\Rightarrow v = - 2$ (rejected), $- {{10} \over 9}$ (accepted)

$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$

$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$

$\therefore$ $18(a + b + c)$

$= 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$

$= 44$"

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