Explanation
We know that,
cosθ=→b1.→b2→|b1||→b2| ...... (1)
Given that,
l+m+n=0
l+m=−n
−(l+m)=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 b1⋅b2.
b1.b2=(0,1,−1).(1,0,−1)
=0+0+1
=1
Therefore,
|b1|=√02+12+(−1)2=√2
|b2|=√02+12+(−1)2=√2
Now, substitute the values in equation (1).
cosθ=→b1.→b2|→b1||→b2|
⇒cosθ=1√2.√2=12
⇒θ=π3
Hence, the angle between the lines is π3.
→r=→a+x(→b−→a)→a=−3ˆi+4ˆj+(−8)ˆk→b=5ˆi+(−6)ˆj+4ˆk→r=−3ˆi+4ˆj+(−8)ˆk+x(8ˆi+(−10)ˆj+12ˆk)=(8x−3)ˆi+(4−10x)ˆj+(12x−8)ˆk
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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