Explanation
We know that,
\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}.}\overrightarrow{{{b}_{2}}}}{\overrightarrow{\left| {{b}_{1}} \right|}\left| {{\overrightarrow{b}}_{2}} \right|} ...... (1)
Given that,
l+m+n=0
l+m=-n
-\left( l+m \right)=n
And, lm=0
So,
Either l=0 or m=0
When, l=0, then
m= -n
And,
(l, m, n)= (0, 1, -1)
When, m=0, then
l= -n
(I, m, n)= (1, 0, -1)
Calculate b_1\cdot b_2.
{{b}_{1}}.{{b}_{2}}=(0,1,-1).(1,0,-1)
=0+0+1
=1
Therefore,
\left| {{b}_{1}} \right|=\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2}
\left| {{b}_{2}} \right|=\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2}
Now, substitute the values in equation (1).
\cos \theta =\dfrac{\overrightarrow{{{b}_{1}}}.\overrightarrow{{{b}_{2}}}}{\left| {{\overrightarrow{b}}_{1}} \right|\left| \overrightarrow{{{b}_{2}}} \right|}
\Rightarrow \cos \theta =\dfrac{1}{\sqrt{2}.\sqrt{2}}=\dfrac{1}{2}
\Rightarrow \theta =\dfrac{\pi }{3}
Hence, the angle between the lines is \dfrac{\pi}{3}.
\overrightarrow { r } =\overrightarrow { a } +x(\overrightarrow { b } -\overrightarrow { a } )\\ \overrightarrow { a } =-3\hat { i } +4\hat { j } +(-8)\hat { k } \\ \overrightarrow { b } =5\hat { i } +(-6)\hat { j } +4\hat { k } \\ \overrightarrow { r } =-3\hat { i } +4\hat { j } +(-8)\hat { k } +x(8\hat { i } +(-10)\hat { j } +12\hat { k } )=(8x-3)\hat { i } +(4-10x)\hat { j } +(12x-8)\hat { k }
On XY plane \hat { k } will be 0
So 12x-8=0 it means x=\dfrac { 2 }{ 3 }
Putting the value of x in component of \hat { i } and \hat { j } we get
8\times \dfrac { 2 }{ 3 } -3 and 4-10\times \dfrac { 2 }{ 3 }
\dfrac { 7 }{ 3 } \ and\ \dfrac { -8 }{ 3 }
So correct optiopn will be option A.
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