The given equation of the line is in vector form: \[ \overrightarrow{r} = \hat{i} + t \hat{j}, \quad t \in \mathbb{R}. \] This represents a line passing through the point \( (1, 0, 0) \) with direction vector \( (0, 1, 0) \).
The symmetric form of a line equation is given by: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, \] where \( (x_1, y_1, z_1) \) is a point on the line and \( (a, b, c) \) is the direction vector.
For this case, the point on the line is \( (1, 0, 0) \), and the direction vector is \( (0, 1, 0) \). Thus, the symmetric form of the equation is: \[ \frac{x - 1}{0} = \frac{y}{1} = \frac{z}{0}. \]
Thus, the correct answer is option (A).
The focus of the parabola \(y^2 + 4y - 8x + 20 = 0\) is at the point:
Let \( S \) denote the set of all subsets of integers containing more than two numbers. A relation \( R \) on \( S \) is defined by:
\[ R = \{ (A, B) : \text{the sets } A \text{ and } B \text{ have at least two numbers in common} \}. \]
Then the relation \( R \) is:
The centre of the hyperbola \(16x^2 - 4y^2 + 64x - 24y - 36 = 0\) is at the point:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
If $ X = A \times B $, $ A = \begin{bmatrix} 1 & 2 \\-1 & 1 \end{bmatrix} $, $ B = \begin{bmatrix} 3 & 6 \\5 & 7 \end{bmatrix} $, find $ x_1 + x_2 $.