The direction cosines of a line satisfy the relations `lambda(l+m)=na n dm n+n l+l m=0.` The value of `lambda,` for which the two lines are perpendicular to each other, is a. `1` b. `2` c. `1//2` d. none of these

Options

A. 1
B. 2
C. `1//2`
D. none of these

Explanation

To solve the problem, we need to find the value of \(\lambda\) for which the two lines represented by the direction cosines are perpendicular to each other. The direction cosines satisfy the following relations:

1. \(\lambda(l + m) = n\) (Equation 1)
2. \(lm + mn + nl = 0\) (Equation 2)

### Step-by-Step Solution:

**Step 1: Rewrite the equations.**

From Equation 1, we can express \(n\) in terms of \(l\) and \(m\):
\[ n = \lambda(l + m) \]

**Step 2: Substitute \(n\) into Equation 2.**

Substituting \(n\) into Equation 2 gives:
\[ lm + m(\lambda(l + m)) + (\lambda(l + m))l = 0 \]

Expanding this:
\[ lm + \lambda(m(l + m)) + \lambda(l(l + m)) = 0 \]
\[ lm + \lambda(ml + m^2 + l^2 + lm) = 0 \]
\[ lm + \lambda(ml + m^2 + l^2 + lm) = 0 \]
\[ lm(1 + \lambda) + \lambda(m^2 + l^2) = 0 \]

**Step 3: Rearranging the equation.**

Rearranging gives us:
\[ lm(1 + \lambda) + \lambda(m^2 + l^2) = 0 \]

**Step 4: Divide by \(m\).**

Dividing the entire equation by \(m\) (assuming \(m \neq 0\)):
\[ l(1 + \lambda) + \lambda\left(\frac{m^2}{m}\right) + \lambda\left(\frac{l^2}{m}\right) = 0 \]
Let \(x = \frac{l}{m}\), then:
\[ x(1 + \lambda) + \lambda(x^2 + 1) = 0 \]

**Step 5: Recognizing the quadratic form.**

This is a quadratic equation in \(x\):
\[ \lambda x^2 + (1 + \lambda)x + \lambda = 0 \]

**Step 6: Finding the product of the roots.**

For a quadratic equation \(ax^2 + bx + c = 0\), the product of the roots is given by \(\frac{c}{a}\). Here:
\[ \text{Product of roots} = \frac{\lambda}{\lambda} = 1 \]

**Step 7: Eliminating \(l\) and \(m\).**

Now, we eliminate \(l\) from the previous equations. Using similar steps as above, we can derive another quadratic equation in terms of \(y = \frac{m}{n}\):
\[ \lambda y^2 - y - 1 = 0 \]

**Step 8: Finding the product of the roots again.**

The product of the roots for this equation is:
\[ \frac{-1}{\lambda} \]

**Step 9: Setting conditions for perpendicular lines.**

For two lines to be perpendicular, the sum of the products of their direction cosines must equal zero:
\[ l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 \]

Using the products of the roots from both quadratic equations, we set:
\[ 1 + 1 - \lambda = 0 \]

**Step 10: Solve for \(\lambda\).**

Solving gives:
\[ 2 - \lambda = 0 \implies \lambda = 2 \]

### Final Answer:
Thus, the value of \(\lambda\) for which the two lines are perpendicular is \(\boxed{2}\).