Equation of the straight lines `3x+2y-z-4=0,4x+y-2z+3=0` in the symmetrical form is …........

Answer

To find the equation of the straight lines given by the equations \(3x + 2y - z - 4 = 0\) and \(4x + y - 2z + 3 = 0\) in symmetrical form, we will follow these steps:

### Step 1: Identify the Direction Cosines
The first step is to express the given equations in terms of direction cosines. We assume the direction cosines of the line are \(L\), \(M\), and \(N\). From the given equations, we can write:

1. \(3L + 2M - N = 0\) (from the first equation)
2. \(4L + M - 2N = 0\) (from the second equation)

### Step 2: Solve the System of Equations
We will solve the system of equations formed by the direction cosines.

From the first equation, we can express \(N\) in terms of \(L\) and \(M\):
\[ N = 3L + 2M \]

Substituting this into the second equation:
\[ 4L + M - 2(3L + 2M) = 0 \]
\[ 4L + M - 6L - 4M = 0 \]
\[ -2L - 3M = 0 \implies 2L + 3M = 0 \implies M = -\frac{2}{3}L \]

### Step 3: Substitute Back to Find \(N\)
Now, substituting \(M\) back into the expression for \(N\):
\[ N = 3L + 2\left(-\frac{2}{3}L\right) = 3L - \frac{4}{3}L = \frac{9L - 4L}{3} = \frac{5L}{3} \]

### Step 4: Normalize the Direction Cosines
Let’s normalize \(L\) to 1:
\[ L = 1, \quad M = -\frac{2}{3}, \quad N = \frac{5}{3} \]

Now we have the direction cosines:
\[ L = 1, \quad M = -\frac{2}{3}, \quad N = \frac{5}{3} \]

### Step 5: Find a Point on the Line
To find a point on the line, we can set \(z = 0\) and solve the equations:
1. \(3x + 2y - 4 = 0\)
2. \(4x + y + 3 = 0\)

From the first equation:
\[ 3x + 2y = 4 \implies 2y = 4 - 3x \implies y = 2 - \frac{3}{2}x \]

Substituting into the second equation:
\[ 4x + (2 - \frac{3}{2}x) + 3 = 0 \]
\[ 4x - \frac{3}{2}x + 5 = 0 \implies \frac{8x - 3x + 10}{2} = 0 \implies 5x + 10 = 0 \implies x = -2 \]

Substituting \(x = -2\) back to find \(y\):
\[ y = 2 - \frac{3}{2}(-2) = 2 + 3 = 5 \]

Thus, a point on the line is \((-2, 5, 0)\).

### Step 6: Write the Symmetrical Form
Using the point \((-2, 5, 0)\) and the direction ratios \(L = 1\), \(M = -\frac{2}{3}\), \(N = \frac{5}{3}\), the symmetrical form of the line is:
\[ \frac{x + 2}{1} = \frac{y - 5}{-\frac{2}{3}} = \frac{z - 0}{\frac{5}{3}} \]

This simplifies to:
\[ x + 2 = -\frac{3}{2}(y - 5) = \frac{3}{5}z \]

### Final Answer
The equation of the straight lines in symmetrical form is:
\[ \frac{x + 2}{1} = \frac{y - 5}{-\frac{2}{3}} = \frac{z}{\frac{5}{3}} \]