Question

Let \( L_1 : \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \) and \( L_2 : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha} \), where \( \alpha \in \mathbb{R} \), be two lines which intersect at the point \( B \). If \( P \) is the foot of the perpendicular from the point \( A(1, 1, -1) \) on \( L_2 \), then the value of \( 26 \alpha (PB)^2 \) is:

Hint:

For perpendicular distance problems, use the parametric equations of the lines and the formula for the distance from a point to a line to find the required distances.

Answer 1

Correct Answer: 216

Solution for the Geometry Problem 

We are given two lines \( L_1 \) and \( L_2 \) with the following equations:

Line \( L_1 \): \[ \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \]

Line \( L_2 \): \[ \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha} \]

The intersection point \( B \) of the two lines is calculated, and the foot of the perpendicular \( P \) from point \( A(1, 1, -1) \) on line \( L_2 \) is also found.

Step 1: Parametric Equations of the Lines

For line \( L_1 \), the parametric equations are:

\[ x = 1 + 3t, \quad y = 1 - t, \quad z = -1 \] where \( t \) is the parameter.

For line \( L_2 \), the parametric equations are:

\[ x = 2 + 2s, \quad y = 0, \quad z = -4 + \alpha s \] where \( s \) is the parameter.

Step 2: Finding the Intersection Point \( B \)

To find the intersection point \( B \), we solve the system of equations by equating the corresponding coordinates from both lines: 1. \( 1 + 3t = 2 + 2s \) 2. \( 1 - t = 0 \) 3. \( -1 = -4 + \alpha s \) Solving these equations: - From the second equation: \( t = 1 \) - Substituting \( t = 1 \) into the first equation: \[ 1 + 3(1) = 2 + 2s \quad \Rightarrow \quad 4 = 2 + 2s \quad \Rightarrow \quad s = 1 \] - Substituting \( s = 1 \) into the third equation: \[ -1 = -4 + \alpha(1) \quad \Rightarrow \quad \alpha = 3 \] So, the intersection point \( B \) is \( B(5, 0, -1) \).

Step 3: Finding the Foot of the Perpendicular \( P \)

The coordinates of the foot of the perpendicular from point \( A(1, 1, -1) \) on line \( L_2 \) are calculated using the perpendicular condition. The vector \( \overrightarrow{AP} \) is perpendicular to the direction vector of \( L_2 \), which is \( (2, 0, 3) \). Solving for the parameter \( s \), we find that \( s = \frac{7}{13} \). Substituting \( s = \frac{7}{13} \) into the parametric equations of \( L_2 \), we get: \[ x_P = \frac{40}{13}, \quad z_P = -\frac{31}{13} \] So, the coordinates of \( P \) are \( P\left( \frac{40}{13}, 0, -\frac{31}{13} \right) \).

Step 4: Finding the Distance \( PB \)

Now, we calculate the distance between \( P \) and \( B(5, 0, -1) \): \[ PB = \sqrt{\left( 5 - \frac{40}{13} \right)^2 + \left( 0 - 0 \right)^2 + \left( -1 + \frac{31}{13} \right)^2} \] Simplifying: \[ 5 - \frac{40}{13} = \frac{25}{13}, \quad -1 + \frac{31}{13} = \frac{18}{13} \] Thus, we get: \[ PB = \sqrt{\left( \frac{25}{13} \right)^2 + \left( \frac{18}{13} \right)^2} = \sqrt{\frac{625}{169} + \frac{324}{169}} = \sqrt{\frac{949}{169}} = \frac{\sqrt{949}}{13} \]

Step 5: Finding \( 26 \alpha (PB)^2 \)

We are asked to find \( 26 \alpha (PB)^2 \). Since \( \alpha = 3 \), we have: \[ 26 \alpha (PB)^2 = 26 \times 3 \times \left( \frac{\sqrt{949}}{13} \right)^2 = 26 \times 3 \times \frac{949}{169} \] Simplifying: \[ 26 \times 3 \times \frac{949}{169} = \frac{26 \times 3 \times 949}{169} = \frac{73962}{169} = 216 \]

Final Answer:

Therefore, the value of \( 26 \alpha (PB)^2 \) is: \[ \boxed{216} \]

Answer 2

Interpret the zero denominators: from \(L_1\) we get \(z=-1\) and the direction vector \(\vec d_1=(3,-1,0)\). A point on \(L_1\) is \(A_1=(1,1,-1)\).

From \(L_2\) we have \(y=0\), a point on \(L_2\) is \(D_0=(2,0,-4)\) and its direction vector is \(\vec d_2=(2,0,\alpha)\).

Find the intersection \(B\). For \(L_1\) put parameter \(s\): \(x=1+3s,\;y=1-s,\;z=-1\). Intersection with \(y=0\) gives \(1-s=0\Rightarrow s=1\Rightarrow x=4\). So \(B=(4,0,-1)\). For \(L_2\) with parameter \(t\): \(x=2+2t,\;y=0,\;z=-4+\alpha t\). Putting \(z=-1\) gives \(\alpha t=3\Rightarrow t=3/\alpha\). Equate \(x\)-coordinates: \(2+2(3/\alpha)=4\Rightarrow 6/\alpha=2\Rightarrow \alpha=3\).

So \(\alpha=3\), \(B=(4,0,-1)\) and \(L_2\) has direction \(\vec d=(2,0,3)\) and point \(D_0=(2,0,-4)\).

Foot of perpendicular \(P\) from \(A(1,1,-1)\) to \(L_2\): write \(P=D_0+t\vec d\). The condition \((A-P)\cdot\vec d=0\) gives \[ t=\frac{(A-D_0)\cdot\vec d}{\vec d\cdot\vec d}. \] Compute \(A-D_0=(-1,1,3)\), \((A-D_0)\cdot\vec d=-1\cdot2+1\cdot0+3\cdot3=7\), and \(\vec d\cdot\vec d=2^2+0^2+3^2=13\). Hence \(t=\dfrac{7}{13}\).

Thus \[ P=D_0+t\vec d=\Big(2+\frac{14}{13},\,0,\, -4+\frac{21}{13}\Big) =\Big(\frac{40}{13},\,0,\,-\frac{31}{13}\Big). \]

Compute \(\overrightarrow{PB}=B-P\): \[ B=\Big(\frac{52}{13},0,-\frac{13}{13}\Big),\quad \overrightarrow{PB}=\Big(\frac{12}{13},0,\frac{18}{13}\Big). \] Hence \[ (PB)^2=\Big(\frac{12}{13}\Big)^2+\Big(\frac{18}{13}\Big)^2=\frac{144+324}{169}=\frac{468}{169}=\frac{36}{13}. \]

Finally \[ 26\alpha\,(PB)^2 = 26\cdot 3 \cdot \frac{36}{13} =\big(2\cdot13\big)\cdot3\cdot\frac{36}{13} =2\cdot3\cdot36=216. \]

Answer

\(26\alpha\,(PB)^2 = 216.\)