Question

Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:

• A. \( \frac{1}{\sqrt{5}} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \frac{1}{\sqrt{3}} \)
• D. \( \frac{1}{2\sqrt{3}} \)

Hint:

For finding the angle between two lines, use the dot product formula and ensure to calculate the direction ratios carefully.

Answer 1

The Correct Option is C

Let the direction ratios of the line \( L \) be \( \hat{i} - \hat{k} \). 
Step 1: The foot of the perpendicular \( N \) lies on the line \( L \), so let \( \overrightarrow{N} = (\lambda, 0, -\lambda) \). 
Step 2: The vector \( \overrightarrow{PN} = (\lambda - 1, 0 - 2, -\lambda + 1) = (\lambda - 1, -2, -\lambda + 1) \). The vector \( \overrightarrow{PQ} \) is parallel to the plane, so its direction ratios are proportional to the normal vector \( \mathbf{n} = \hat{i} + \hat{j} + 2\hat{k} \). 
Step 3: Now, calculate the cosine of the angle between the two lines \( PN \) and \( PQ \) using the dot product formula: \[ \cos \alpha = \frac{\overrightarrow{PN} \cdot \overrightarrow{PQ}}{|\overrightarrow{PN}| |\overrightarrow{PQ}|} \] After calculation, we find: \[ \cos \alpha = \frac{1}{\sqrt{3}} \]

Answer 2

Step 1: Understanding the Given Information
We are given the following details:
- Point \( P(1, 2, -1) \)
- The straight line \( L \) is represented as \( \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \), which can be written in parametric form as:
\[ x = t, \, y = 0, \, z = -t \] where \( t \) is the parameter.
- We need to find the foot of the perpendicular from point \( P \) to the line \( L \), denoted as \( N \).
- A line is drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \), and it intersects the line \( L \) at point \( Q \). We are asked to find the acute angle \( \alpha \) between the lines \( PN \) and \( PQ \), and specifically, the value of \( \cos \alpha \).
Step 2: Foot of Perpendicular \( N \)
To find the foot of the perpendicular \( N \) from point \( P(1, 2, -1) \) to the line \( L \), we need the direction vector of the line \( L \), which is \( \vec{d} = (1, 0, -1) \). Let the position vector of point \( N \) be \( (t, 0, -t) \), where \( t \) is the parameter of the line \( L \).
The vector from \( P(1, 2, -1) \) to \( N(t, 0, -t) \) is:
\[ \overrightarrow{PN} = (t - 1, 0 - 2, -t + 1) = (t - 1, -2, -t + 1) \] For the vector \( \overrightarrow{PN} \) to be perpendicular to the direction vector \( \vec{d} = (1, 0, -1) \), their dot product must be zero:
\[ (t - 1) \cdot 1 + (-2) \cdot 0 + (-t + 1) \cdot (-1) = 0 \] Simplifying this equation:
\[ t - 1 + t - 1 = 0 \quad \Rightarrow \quad 2t - 2 = 0 \quad \Rightarrow \quad t = 1 \] Thus, the foot of the perpendicular \( N \) is the point \( (1, 0, -1) \).
Step 3: Line Parallel to the Plane \( x + y + 2z = 0 \)
The direction vector of the plane \( x + y + 2z = 0 \) is the normal vector \( \vec{n} = (1, 1, 2) \). A line parallel to the plane will have a direction vector that is perpendicular to the normal vector \( \vec{n} \). Since the line passes through \( P(1, 2, -1) \), its direction vector will be perpendicular to \( \vec{n} \), and hence it can be obtained by solving the following system of equations.
Let the direction vector of the line \( PQ \) be \( \vec{v} = (a, b, c) \). The condition for the line to be parallel to the plane is:
\[ \vec{n} \cdot \vec{v} = 0 \quad \Rightarrow \quad (1, 1, 2) \cdot (a, b, c) = a + b + 2c = 0 \] This is the equation of the direction vector of line \( PQ \).
Step 4: Finding the Acute Angle Between \( PN \) and \( PQ \)
The acute angle \( \alpha \) between the two lines is given by the formula:
\[ \cos \alpha = \frac{\overrightarrow{PN} \cdot \overrightarrow{PQ}}{|\overrightarrow{PN}| |\overrightarrow{PQ}|} \] where \( \overrightarrow{PN} = (0, -2, 0) \) and \( \overrightarrow{PQ} \) is the direction vector of the line from \( P \) to \( Q \).
After solving for the direction vectors and calculating the cosine of the angle, we find that:
\[ \cos \alpha = \frac{1}{\sqrt{3}} \] Step 5: Conclusion
The value of \( \cos \alpha \) is \( \frac{1}{\sqrt{3}} \).