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The direction cosines of two lines are c...

The direction cosines of two lines are connected by relation `l+m+n=0` and 4l is the harmonic mean between m and n.
Then,

A

`(l_(1)/l_(2))+m_(1)/m_(2)+n_(1)/n_(2)=-3//2`

B

`l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=-1/2`

C

`l_(1)m_(1)n_(1)+l_(2)m_(2)n_(2)=-sqrt(6)//9`

D

`(l_(1)+l_(2))(m_(1)+m_(2))(n_(1)+n_(2))=sqrt(6)/18`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`l+m+n=0`, `4l=(2mn)/(m+n)`
`rArr 2lm-mn+2ln=0`
By eliminating n, we get
`2(l/m)^(2)-l/m-1=0`
`therefore l_(1)/m_(1)=1, l_(2)/m_(2)=-1/2`
On solving, we get
`l_(1)m_(2),n_(2)0=(1/sqrt(6),1/sqrt(6),-2/sqrt(6))`
and `(l_(2),m_(2),n_(2))=(1/sqrt(6),-2/sqrt(6),1/sqrt(6))`
`l_(1)/l_(2) + m_(1)/m_(2)+n_(1)/n_(2)=1-1/2-2=-3/2`
`l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=1/6-2/6-2/6=-1/2`
`l_(1)m_(1)n_(1) + l_(2)m_(2)n_(2)=-2/(6sqrt(6))-2/(6sqrt(6))=-2/(3sqrt(6))=(-2sqrt(6))/18=-sqrt(16)/9`
`therefore (l_(1)+l_(2))(m_(1)+m_(2))(n_(1)+n_(2))=2/sqrt(6) xx -1/sqrt(6) xx -1/sqrt(6)`
`=2/(6sqrt(6)) = sqrt(6)/18`
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