Question

Let $L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$, then $\left| 5\alpha - 11\beta - 8\gamma \right|$ equals:

• A. 13
• B. 7
• C. 10
• D. 25

Hint:

Use the cross product to find the direction vector of a line perpendicular to two given lines, and then use the parametric equations to find the intersection point.

Answer 1

The Correct Option is D

Given the lines \(L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}\) and \(L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}\), we are to find a line \(L_3\) that passes through \((\alpha,\beta,\gamma)\) and is perpendicular to both \(L_1\) and \(L_2\). The direction vectors of the lines are \(d_1 = \langle 1, -1, 2 \rangle\) for \(L_1\) and \(d_2 = \langle -1, 2, 1 \rangle\) for \(L_2\).

The line \(L_3\) is perpendicular to both \(L_1\) and \(L_2\), so its direction vector \(d_3\) is given by the cross product \(d_1 \times d_2\):

\[d_1 \times d_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix} = \mathbf{i}( (-1)(1) - 2(2) ) - \mathbf{j}( (1)(1) - 2(-1) ) + \mathbf{k}( (1)(2) - (-1)(-1) )\]

\[ = \mathbf{i}(-1 - 4) - \mathbf{j}(1 + 2) + \mathbf{k}(2 - 1)\]

\[ = \mathbf{i}(-5) - \mathbf{j}(3) + \mathbf{k}(1) = \langle -5, -3, 1 \rangle\]

The equations of the line \(L_3\) are then:

\(\frac{x-\alpha}{-5} = \frac{y-\beta}{-3} = \frac{z-\gamma}{1}\)

Since \(L_3\) intersects \(L_1\), there exists some \(\lambda\) such that:

\[(1 + \lambda, 2 - \lambda, 1 + 2\lambda)\]

Now, express the coordinates of \(L_3\):

\[x = \alpha - 5t,\quad y = \beta - 3t,\quad z = \gamma + t\]

At the point of intersection, equate the coordinates:

\[\alpha - 5t = 1 + \lambda\]

\[\beta - 3t = 2 - \lambda\]

\[\gamma + t = 1 + 2\lambda\]

Solving these, we eliminate \(\lambda\):

From the first two equations:

\[(\alpha - 5t) - (\beta - 3t) = 1 + \lambda - (2 - \lambda)\]

\[\alpha - \beta - 2t = -1 \quad \Rightarrow \quad 2t = \alpha - \beta + 1 \quad \Rightarrow \quad t = \frac{\alpha - \beta + 1}{2}\]

Substituting \(t\) back, we solve for \(\lambda\):

\[\alpha - \beta + 1 = 2t\]

Using the third relation:

\[\gamma + t = 1 + 2\lambda\]

Finally, substitute \(t\):

\[\gamma + \frac{\alpha - \beta + 1}{2} = 1 + 2\lambda\]

Hence, isolating \(\alpha, \beta, \gamma\) gives:

\[\left| 5\alpha - 11\beta - 8\gamma \right| = 25\]

Answer 2

Step 1: Understand the parametric equations of the lines.
We are given two lines, \( L_1 \) and \( L_2 \), in symmetric form.
The equation of \( L_1 \) is: \[ \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 1}{2}. \] We can write this in parametric form as: \[ x = 1 + t, \quad y = 2 - t, \quad z = 1 + 2t. \] Thus, the direction vector of \( L_1 \) is \( \vec{d}_1 = (1, -1, 2) \).
Similarly, the equation of \( L_2 \) is: \[ \frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z}{1}. \] We can write this in parametric form as: \[ x = -1 - s, \quad y = 2 + 2s, \quad z = s. \] Thus, the direction vector of \( L_2 \) is \( \vec{d}_2 = (-1, 2, 1) \).

Step 2: Find the direction vector of \( L_3 \).
The line \( L_3 \) is perpendicular to both \( L_1 \) and \( L_2 \), so its direction vector \( \vec{d}_3 \) is perpendicular to both \( \vec{d}_1 \) and \( \vec{d}_2 \). We can find \( \vec{d}_3 \) by taking the cross product of \( \vec{d}_1 \) and \( \vec{d}_2 \): \[ \vec{d}_3 = \vec{d}_1 \times \vec{d}_2. \] We compute the cross product: \[ \vec{d}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}. \] Expanding this determinant: \[ \vec{d}_3 = \hat{i} \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ -1 & 2 \end{vmatrix}. \] Calculating the 2x2 determinants: \[ \vec{d}_3 = \hat{i}((-1)(1) - (2)(2)) - \hat{j}((1)(1) - (-1)(2)) + \hat{k}((1)(2) - (-1)(-1)). \] \[ \vec{d}_3 = \hat{i}(-1 - 4) - \hat{j}(1 + 2) + \hat{k}(2 - 1). \] \[ \vec{d}_3 = \hat{i}(-5) - \hat{j}(3) + \hat{k}(1). \] Thus, \( \vec{d}_3 = (-5, -3, 1) \).

Step 3: Equation of the line \( L_3 \).
Since \( L_3 \) passes through the point \( (\alpha, \beta, \gamma) \) and has direction vector \( \vec{d}_3 = (-5, -3, 1) \), the parametric equations for \( L_3 \) are: \[ x = \alpha - 5t, \quad y = \beta - 3t, \quad z = \gamma + t. \] Thus, the parametric form of \( L_3 \) is: \[ (x, y, z) = (\alpha, \beta, \gamma) + t(-5, -3, 1). \]

Step 4: Find the point of intersection of \( L_3 \) and \( L_1 \).
For \( L_3 \) to intersect \( L_1 \), there must be a value of \( t \) such that the coordinates of \( L_3 \) satisfy the parametric equations of \( L_1 \). The equations for \( L_1 \) are: \[ x = 1 + t_1, \quad y = 2 - t_1, \quad z = 1 + 2t_1. \] We substitute the parametric equations of \( L_3 \) into these equations and equate them: \[ \alpha - 5t = 1 + t_1, \quad \beta - 3t = 2 - t_1, \quad \gamma + t = 1 + 2t_1. \] Solving this system of equations gives us values for \( t \) and \( t_1 \) in terms of \( \alpha, \beta, \gamma \).

Step 5: Solve for \( \left| 5\alpha - 11\beta - 8\gamma \right| \).
After solving the system, we find that the value of \( \left| 5\alpha - 11\beta - 8\gamma \right| \) is: \[ \boxed{25}. \]