Question

If the line with direction ratios \[ (1, a, \beta) \] is perpendicular to the line with direction ratios \[ (-1,2,1) \] and parallel to the line with direction ratios \[ (\alpha,1,\beta), \] then \( (\alpha, \beta) \) is:

• A. \( (-1,-1) \)
• B. \( (1,-1) \)
• C. \( (1,3) \)
• D. \( (1,1) \)

Hint:

For perpendicularity in 3D, use the dot product condition, and for parallelism, equate the ratios of direction cosines.

Answer 1

The Correct Option is B

Step 1: Using perpendicularity condition Two lines are perpendicular if: \[ \alpha_1 \alpha_2 + \beta_1 \beta_2 + \gamma_1 \gamma_2 = 0. \] Solving with the given ratios: \[ 1(-1) + a(2) + \beta(1) = 0. \] Step 2: Using parallel condition Since the lines are parallel, \[ \alpha = 1, \quad \beta = -1. \]

Answer 2

Given:
- A line with direction ratios \((1, a, \beta)\)
- It is perpendicular to the line with direction ratios \((-1, 2, 1)\)
- It is parallel to the line with direction ratios \((\alpha, 1, \beta)\)

We need to find \( (\alpha, \beta) \).

Step 1: Since the line \((1, a, \beta)\) is perpendicular to \((-1, 2, 1)\), their direction ratios satisfy:
\[ 1 \times (-1) + a \times 2 + \beta \times 1 = 0 \] \[ -1 + 2a + \beta = 0 \] \[ 2a + \beta = 1 \quad \Rightarrow \quad \beta = 1 - 2a \]

Step 2: Since the line \((1, a, \beta)\) is parallel to \((\alpha, 1, \beta)\), their direction ratios are proportional:
\[ \frac{1}{\alpha} = \frac{a}{1} = \frac{\beta}{\beta} \] Note that \(\frac{\beta}{\beta} = 1\) (assuming \(\beta \neq 0\)).
So:
\[ a = 1 \quad \text{and} \quad \frac{1}{\alpha} = 1 \implies \alpha = 1 \]

Step 3: Substitute \( a = 1 \) into \(\beta = 1 - 2a\):
\[ \beta = 1 - 2 \times 1 = 1 - 2 = -1 \]

Therefore,
\[ (\alpha, \beta) = (1, -1) \]