Respected Sir, please help me through the following problem:

Find the equations of the two lines through the origin which intersect the line(x-3)/2=(y-3)/1=z/1 at an angle of 60 degrees.

To solve the problem of finding the equations of the two lines through the origin that intersect the given line at an angle of 60 degrees, we need to break it down into manageable steps. First, let's analyze the line represented by the equation \((x-3)/2=(y-3)/1=z/1\). This can be rewritten in parametric form, which will help us understand its direction vector.

Step 1: Parametric Form of the Given Line

The equation \((x-3)/2=(y-3)/1=z/1\) can be expressed in parametric form as follows:

• x = 2t + 3
• y = t + 3
• z = t

Here, \(t\) is a parameter. The direction vector of this line, derived from the coefficients of \(t\), is v = (2, 1, 1).

Step 2: Understanding the Angle Between Lines

To find the lines through the origin that intersect this line at a 60-degree angle, we need to use the concept of the dot product. The angle \(\theta\) between two vectors can be found using the formula:

cos(θ) = (u · v) / (|u| |v|)

Where \(u\) is the direction vector of the line we are looking for, and \(v\) is the direction vector of the given line. For our case, we want \(\theta = 60^\circ\), which means:

cos(60°) = 1/2

Step 3: Setting Up the Equation

Let the direction vector of the lines through the origin be u = (a, b, c). The dot product \(u · v\) is:

u · v = 2a + b + c

The magnitude of \(u\) is \(|u| = √(a² + b² + c²)\) and the magnitude of \(v\) is \(|v| = √(2² + 1² + 1²) = √6\).

Substituting these into the cosine formula gives us:

(2a + b + c) / (√(a² + b² + c²) * √6) = 1/2

Step 4: Rearranging the Equation

Cross-multiplying leads to:

2(2a + b + c) = √6 * √(a² + b² + c²)

Squaring both sides to eliminate the square root gives:

4(2a + b + c)² = 6(a² + b² + c²)

Step 5: Finding the Direction Vectors

Now, we need to find specific values for \(a\), \(b\), and \(c\) that satisfy this equation. One way to approach this is to assume values for two variables and solve for the third. For example, let’s set \(c = 0\) to simplify our calculations:

Then the equation reduces to:

4(2a + b)² = 6(a² + b²)

Expanding and rearranging gives:

8a² + 8ab + 4b² = 6a² + 6b²

Which simplifies to:

2a² + 8ab - 2b² = 0

Factoring out 2 gives:

a² + 4ab - b² = 0

Step 6: Solving the Quadratic Equation

This is a quadratic in terms of \(a\):

a² + 4ab - b² = 0

Using the quadratic formula \(a = \frac{-B \pm \sqrt{B² - 4AC}}{2A}\), where \(A = 1\), \(B = 4b\), and \(C = -b²\), we find:

a = \frac{-4b \pm \sqrt{(4b)² - 4(1)(-b²)}}{2(1)} = \frac{-4b \pm \sqrt{16b² + 4b²}}{2} = \frac{-4b \pm \sqrt{20b²}}{2}

This simplifies to:

a = -2b \pm b\sqrt{5}

Step 7: Finding the Direction Vectors

Thus, we have two solutions for \(a\):

• a₁ = (-2 + \sqrt{5})b
• a₂ = (-2 - \sqrt{5})b

Choosing \(b = 1\) for simplicity, we can find the two direction vectors:

• Line 1: (−2 + √5, 1, 0)
• Line 2: (−2 - √5, 1, 0)

Step 8: Writing the Equations of the Lines

Finally, the equations of the two lines through the origin can be expressed in parametric form as:

• Line 1: x = (−2 + √5)t, y = t, z = 0
• Line 2: x = (−2 - √5)t, y = t, z = 0

These equations represent the two lines that intersect the given line at an angle of 60 degrees. If you have any further questions or need clarification on any step, feel free to ask!