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

Consider two 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 need to determine the equation of a line \(L_3\) that passes through the point \((\alpha,\beta,\gamma)\) and is orthogonal to both \(L_1\) and \(L_2\). The direction vectors for \(L_1\) and \(L_2\) are \(d_1 = \langle 1, -1, 2 \rangle\) and \(d_2 = \langle -1, 2, 1 \rangle\), respectively.

Since \(L_3\) is perpendicular to both \(L_1\) and \(L_2\), its direction vector \(d_3\) is the cross product of \(d_1\) and \(d_2\):

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

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

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

The parametric equations for line \(L_3\) are:

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

As \(L_3\) intersects \(L_1\), there exists a parameter \(\lambda\) such that a point on \(L_1\) can be represented as:

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

The parametric representation of \(L_3\) is:

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

Equating the coordinates at the point of intersection yields:

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

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

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

Eliminating \(\lambda\) from these equations:

Subtracting the second equation from the first:

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

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

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

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

Substitute this expression for \(t\) into the first two equations to solve for \(\lambda\):

From \( \alpha - 5t = 1 + \lambda \), we get \( \lambda = \alpha - 5t - 1 \).

From \( \beta - 3t = 2 - \lambda \), we get \( \lambda = 2 - \beta + 3t \).

Equating these gives \( \alpha - 5t - 1 = 2 - \beta + 3t \), which simplifies to \( \alpha + \beta - 3 = 8t \).

Using the third equation, \( \gamma + t = 1 + 2\lambda \). Substitute \( \lambda = \alpha - 5t - 1 \):

\[\gamma + t = 1 + 2(\alpha - 5t - 1)\]

\[\gamma + t = 1 + 2\alpha - 10t - 2\]

\[\gamma + t = 2\alpha - 10t - 1\]

\[11t = 2\alpha - \gamma - 1\]

Now substitute \(t = \frac{\alpha - \beta + 1}{2}\) into \(11t = 2\alpha - \gamma - 1\):

\[11\left(\frac{\alpha - \beta + 1}{2}\right) = 2\alpha - \gamma - 1\]

\[11(\alpha - \beta + 1) = 2(2\alpha - \gamma - 1)\]

\[11\alpha - 11\beta + 11 = 4\alpha - 2\gamma - 2\]

\[7\alpha - 11\beta + 2\gamma + 13 = 0\]

This implies the condition relating \(\alpha, \beta, \gamma\) for the intersection is \(7\alpha - 11\beta + 2\gamma + 13 = 0\). The absolute value given in the original text seems to be derived from a different calculation or context not fully explained here.