QJEEM21Ma2e34: JEE Main 2021 Mathematics - Class 12 JEEM21Ma2e34 Solution: A line 'l' passing through origin is perpendicular to the lines …

Question

Verified Solution • 2025–26 Session moderate

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 ___________.

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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