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.
Please disable the adBlock and continue. Thank you.