Question

The equation of the line through the point (1, 1, 1) and parallel to the line joining the points (-2, 2, 0) and (-1, 1, 1) is

• A. \(\frac{x-1}{-3}=\frac{y-1}{-1}=z-1\)
• B. 1-x=1+y=1-z
• C. x+1=-(y-1)=z-1
• D. \(\frac{x-1}{-1}=\frac{y+1}{2}=\frac{z-1}{1}\)
• E. x+2=y-2=z

Answer 1

The Correct Option is B

The equation of a line in space passing through a point \( P(x_1, y_1, z_1) \) and parallel to a vector \( \mathbf{d} = (a, b, c) \) is given by: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \] We are given the point \( (1, -1, 1) \) and the line joining \( (-2, 2, 0) \) and \( (-1, 1, 1) \).
Step 1: Find the direction vector of the line joining \( (-2, 2, 0) \) and \( (-1, 1, 1) \) The direction vector is: \[ \mathbf{d} = \langle -1 - (-2), 1 - 2, 1 - 0 \rangle = \langle 1, -1, 1 \rangle \]
Step 2: Use the formula for the equation of the line The equation of the line through \( (1, -1, 1) \) and parallel to the direction vector \( \mathbf{d} = \langle 1, -1, 1 \rangle \) is: \[ \frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z - 1}{1} \] Simplifying this, we get: \[ 1 - x = 1 + y = 1 - z \] 

The correct option is (B) : \(1-x=1+y=1-z\)

Answer 2

The line passes through the point (1, 1, 1). The line is parallel to the line joining the points (-2, 2, 0) and (-1, 1, 1). The direction vector of the line joining (-2, 2, 0) and (-1, 1, 1) is \((-1 - (-2), 1 - 2, 1 - 0) = (1, -1, 1)\).

Since the line is parallel, it has the same direction vector, so the direction vector of the line we want is (1, -1, 1).

The equation of a line passing through a point \((x_0, y_0, z_0)\) with direction vector \((a, b, c)\) is given by \(\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\).

In our case, \((x_0, y_0, z_0) = (1, 1, 1)\) and \((a, b, c) = (1, -1, 1)\). Therefore, the equation of the line is:

\(\frac{x - 1}{1} = \frac{y - 1}{-1} = \frac{z - 1}{1}\)

This can be written as:

\(x - 1 = -(y - 1) = z - 1\)

\(x - 1 = 1 - y = z - 1\)

\(1-x=y-1 = 1-z\)

1-x=-(z-1)=-y+1 1-x=1-y=1-z

Therefore, the equation of the line is \(1-x = y-1 = 1-z\).