Question

the direction cosines of two lines are determined by the relation l-5m+3n=0 and 7l^2+5m^2-3n^2=0, find them

Answer 1

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here is ur answer...

Solution:

The given relations are

l – 5m + 3n = 0 ⇒ l = 5m – 3n   ……(1)

and 7l2 + 5m2 – 3n2 = 0               ……(2)

Putting the value of l from (1) in (2), we get 

7(5m – 3n)2 + 5m2 – 3n2 = 0

or, 180m2 – 210mn + 60n2 = 0 

or, (2m – n)(3m – 2n) = 0

∴ m/n = 1/2 or 2/3

when m/n = 1/2 i.e. n = 2m

∴ l = 5m – 3n = –m or 1/m = –1

Hence, we have l/–1 = m/1 = n/2 = √ (l2 + m2 + n2) / √ {(–1)2 + l2 + 22} = 1/√6

So, direction cosines of one line are –1/√6, 1/√6, 2/√6

Again when m/n = 2/3 or n = 3m/2

∴ l = 5m – 3.3m/2 = m/2 or 1/m = 1/2

Thus, m/n = 2/3 and l/m = 1/2 giving l/1 = m/2 = n/3 = 1/√12 + 22 + 32 = 1/√14

∴ The direction cosines of the other line are 1/√14, 2/√14, 3/√14.

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Answer 2

Answer:

In order to compute the values of l, m and n from the given relations, we shall first solve these equations.  

We have l - 5m + 3n = 0 ⇒ l =5m-3n

Substituting this value in the second equation we have

7(5m-3n) 2 + 5m2 – 3n2 = 0

Hence, 30(2m-n)(3m-2n) = 0 i.e. 2m = n and 3m = 2n.

Therefore, m/1 = n/2 = (5m-3n)/5-2.3 = l/(-1) = 1/√6

and m/2 = n/3 = (5m-3n)/ (5.2 – 3.3) = l/1 = 1/√14

Hence, the required dc’s of the line are -1/√6, 1/√6, 2/√6 and 1/√14, 2/√14, 3/√14.

Step-by-step explanation: