Step 1: Parametric equation of the line. The given line can be expressed in parametric form as: \[ x = t, \quad y = 1 + 2t, \quad z = 2 + 3t, \] where \(t\) is a parameter.
Step 2: Midpoint of \(P\) and \(P'\). The image \(P'(1, 0, 7)\) of \(P(x, y, z)\) with respect to the given line implies that the midpoint \(M\) of \(P\) and \(P'\) lies on the line.
The coordinates of the midpoint \(M\) are: \[ M = \left(\frac{x + 1}{2}, \frac{y + 0}{2}, \frac{z + 7}{2}\right). \]
Step 3: Condition for \(M\) lying on the line. Since \(M\) lies on the line, its coordinates must satisfy the parametric equations of the line. Thus: \[ \frac{x + 1}{2} = t, \quad \frac{y}{2} = 1 + 2t, \quad \frac{z + 7}{2} = 2 + 3t. \]
Step 4: Solve for \(t\). From the first equation: \[ t = \frac{x + 1}{2}. \] Substitute \(t\) into the second equation: \[ \frac{y}{2} = 1 + 2\left(\frac{x + 1}{2}\right). \]
Simplify: \[ \frac{y}{2} = 1 + x + 1 \implies \frac{y}{2} = x + 2 \implies y = 2x + 4. \tag{1} \]
Substitute \(t\) into the third equation: \[ \frac{z + 7}{2} = 2 + 3\left(\frac{x + 1}{2}\right). \]
Simplify: \[ \frac{z + 7}{2} = 2 + \frac{3x + 3}{2} \implies z + 7 = 4 + 3x + 3 \implies z = 3x. \tag{2} \]
Step 5: Use the coordinates of \(P'\) to find \(P\). From the midpoint condition: \[ \frac{x + 1}{2} = t, \quad \frac{y + 0}{2} = 1 + 2t, \quad \frac{z + 7}{2} = 2 + 3t. \]
Substitute \(t = \frac{x + 1}{2}\): \[ \frac{y + 0}{2} = 1 + 2\left(\frac{x + 1}{2}\right). \]
Simplify: \[ \frac{y}{2} = 1 + x + 1 \implies y = 2x + 4. \tag{3} \] Similarly: \[ \frac{z + 7}{2} = 2 + 3\left(\frac{x + 1}{2}\right). \] Simplify: \[ \frac{z + 7}{2} = 2 + \frac{3x + 3}{2} \implies z = 3x. \tag{4} \]
Step 6: Solve for \(P\). We now solve for the coordinates of \(P(x, y, z)\).
Recall the midpoint conditions and verify: From the midpoint: \[ M = \left(\frac{x + 1}{2}, \frac{y}{2}, \frac{z + 7}{2}\right), \] and since \(M\) lies on the line, its coordinates must satisfy: \[ x = t, \quad y = 1 + 2t, \quad z = 2 + 3t. \]
Using the given image \(P'(1, 0, 7)\) and midpoint relationships: \[ \frac{x + 1}{2} = t, \quad \frac{y}{2} = 1 + 2t, \quad \frac{z + 7}{2} = 2 + 3t. \]
Substitute \(t = \frac{x + 1}{2}\) into the second and third equations: 1.
From the second equation: \[ \frac{y}{2} = 1 + 2\left(\frac{x + 1}{2}\right). \] Simplify: \[ \frac{y}{2} = 1 + x + 1 \implies y = 2x + 4. \] 2. From the third equation: \[ \frac{z + 7}{2} = 2 + 3\left(\frac{x + 1}{2}\right). \]
Simplify: \[ \frac{z + 7}{2} = 2 + \frac{3x + 3}{2} \implies z + 7 = 4 + 3x + 3 \implies z = 3x. \] Thus, the coordinates of \(P(x, y, z)\) are determined as: \[ P(x, y, z) = (0, 4, 0). \]
Final Answer: The coordinates of \(P\) are: \[ P(x, y, z) = (0, 4, 0). \]
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Assertion (A): A line in space cannot be drawn perpendicular to \( x \), \( y \), and \( z \) axes simultaneously.
Reason (R): For any line making angles \( \alpha, \beta, \gamma \) with the positive directions of \( x \), \( y \), and \( z \) axes respectively, \[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1. \]