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:

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