P_1:-> r . (3hat i - 5hat j + hat k) = 7 P_2:-> r . (lambda hat i + hat j - 3hat k) = 9 Let angle between P_(1) and P_(2) is theta Then cos theta = 3lambda - 5 - 3 / sqrt 35 sqrt lambda ^2 + 10 But sin theta = 2sqrt 6 / 5 Therefore (3lambda - 8)^2 / 35(lambda ^2 + 10) = 1 - 24 / 25 => 5(9lambda ^2 + 64 - 48lambda ) = 7lambda ^2 + 70 => 38lambda ^2 - 240lambda + 250 = 0 => 19lambda ^2 - 120lambda + 125 = 0 => (19lambda - 25)(lambda - 5) = 0 Therefore lambda _1 = 25 / 19,lambda _2 = 5 So, point (50, 50, 2) Therefore d = |150 - 250 + 2 - 7| / sqrt 35 = 315

If lambda1 andlt lambda2 are two values of lambda such ... | null (Mathematics)

{P_1}:\overrightarrow r \,.\,(3\widehat i - 5\widehat j + \widehat k) = 7

{P_2}:\overrightarrow r \,.\,(\lambda \widehat i + \widehat j - 3\widehat k) = 9

Let angle between P₁ and P₂ is $\theta$

Then

\cos \theta = {{3\lambda - 5 - 3} \over {\sqrt {35} \sqrt {{\lambda ^2} + 10} }}

But $\sin \theta = {{2\sqrt 6 } \over 5}$

\therefore {{{{(3\lambda - 8)}^2}} \over {35({\lambda ^2} + 10)}} = 1 - {{24} \over {25}}

\Rightarrow 5(9{\lambda ^2} + 64 - 48\lambda ) = 7{\lambda ^2} + 70

\Rightarrow 38{\lambda ^2} - 240\lambda + 250 = 0

\Rightarrow 19{\lambda ^2} - 120\lambda + 125 = 0

\Rightarrow (19\lambda - 25)(\lambda - 5) = 0

\therefore {\lambda _1} = {{25} \over {19}},{\lambda _2} = 5

So, point (50, 50, 2)

\therefore d = {{|150 - 250 + 2 - 7|} \over {\sqrt {35} }} = 315

Explanation

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