Question

A straight line through the point (1,-1,0) meets the line $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z-1}{-1}$ at right angle. Its equation is ________.

• A. $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{2}$
• B. $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z}{4}$
• C. $\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z}{6}$
• D. $\frac{x-1}{-1}=\frac{y-1}{-1}=\frac{z}{3}$
• E. $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{-2}$

Hint:

Use the dot product of direction vectors for perpendicular lines.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
We need to find the equation of a line \(L_1\) that passes through a given point \(P(1, -1, 0)\) and is perpendicular to another given line \(L_2\). The line \(L_1\) also intersects \(L_2\). The intersection point is the foot of the perpendicular from point P to the line \(L_2\).
Step 2: Key Formula or Approach
Let \(L_2\) be the line \(\frac{x-1}{1} = \frac{y+1}{1} = \frac{z-1}{-1} = \lambda\).
1. Find the coordinates of a general point \(Q\) on the line \(L_2\) in terms of \(\lambda\).
2. Find the direction vector \(\vec{PQ}\) of the line \(L_1\).
3. The direction vector of \(L_2\) is \(\vec{d_2} = \langle 1, 1, -1 \rangle\).
4. Since \(L_1\) is perpendicular to \(L_2\), their direction vectors must be orthogonal. Use the dot product condition: \(\vec{PQ} \cdot \vec{d_2} = 0\).
5. Solve for \(\lambda\) to find the specific point \(Q\) (the foot of the perpendicular).
6. The required line \(L_1\) passes through P and Q. Its direction vector is \(\vec{PQ}\). Write its equation.
Step 3: Detailed Explanation
1. General point on \(L_2\).
Let \(Q\) be a point on the line \(L_2\). Its coordinates can be written as:
\(x = \lambda + 1\)
\(y = \lambda - 1\)
\(z = -\lambda + 1\)
So, \(Q = (\lambda+1, \lambda-1, -\lambda+1)\).
2. Direction vector \(\vec{PQ}\).
The given point is \(P = (1, -1, 0)\).
\(\vec{PQ} = Q - P = \langle (\lambda+1)-1, (\lambda-1)-(-1), (-\lambda+1)-0 \rangle\)
\(\vec{PQ} = \langle \lambda, \lambda, -\lambda+1 \rangle\). This is the direction vector of our required line \(L_1\).
3. Apply the perpendicularity condition.
The direction vector of line \(L_2\) is \(\vec{d_2} = \langle 1, 1, -1 \rangle\).
Since \(L_1 \perp L_2\), their direction vectors are orthogonal.
\(\vec{PQ} \cdot \vec{d_2} = 0\)
\[ (\lambda)(1) + (\lambda)(1) + (-\lambda+1)(-1) = 0 \] \[ \lambda + \lambda + \lambda - 1 = 0 \] \[ 3\lambda - 1 = 0 \] \[ \lambda = \frac{1}{3} \] 4. Find the direction vector of \(L_1\).
Now substitute \(\lambda = 1/3\) back into the expression for \(\vec{PQ}\):
\(\vec{PQ} = \langle \frac{1}{3}, \frac{1}{3}, -\frac{1}{3}+1 \rangle = \langle \frac{1}{3}, \frac{1}{3}, \frac{2}{3} \rangle\).
This is the direction vector for \(L_1\). Any scalar multiple of this vector is also a valid direction vector. Let's multiply by 3 to get rid of the fractions:
Direction vector \(\vec{d_1} = 3 \cdot \vec{PQ} = \langle 1, 1, 2 \rangle\).
5. Write the equation of line \(L_1\).
The line \(L_1\) passes through the point \(P(1, -1, 0)\) and has direction vector \(\vec{d_1} = \langle 1, 1, 2 \rangle\).
The Cartesian equation is:
\[ \frac{x-1}{1} = \frac{y-(-1)}{1} = \frac{z-0}{2} \] \[ \frac{x-1}{1} = \frac{y+1}{1} = \frac{z}{2} \] This matches option (A).
Step 4: Final Answer
The equation of the line is \(\frac{x-1}{1} = \frac{y+1}{1} = \frac{z}{2}\).