Let $\mathbf{L}_1: \frac{x-1}{1} = \frac{y-2}{1} = \frac{z-1}{2}$ and $\mathbf{L}_2: \frac{x+1}{1} = \frac{y-2}{2} = \frac{z}{4}$ 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 $|5\alpha - 11\beta - 8\gamma|$ equals:

Step 1: Find the direction vector of L3

First, we identify the direction vectors of lines $L_1$ and $L_2$. For $L_1$, the direction vector is $\vec{d}_{1} = \hat{i} + \hat{j} + 2\hat{k}$. For $L_2$, the direction vector is $\vec{d}_{2} = \hat{i} + 2\hat{j} + 4\hat{k}$. Since line $L_3$ is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d}_{3}$ must be parallel to the cross product of $\vec{d}_{1}$ and $\vec{d}_{2}$.

Step 2: Calculate the cross product

We calculate the cross product of $\vec{d}_{1}$ and $\vec{d}_{2}$ to find the direction vector $\vec{d}_{3}$. This results in $0\hat{i} - 2\hat{j} + \hat{k}$. So, we can take $\vec{d}_{3} = -2\hat{j} + \hat{k}$.

Step 3: Represent points on L1 and L3

Any point on line $L_1$ can be represented as $P_1(t) = (1+t, 2+t, 1+2t)$ for some scalar $t$. Line $L_3$ passes through $(\alpha, \beta, \gamma)$ and has direction vector $(0, -2, 1)$. So, any point on $L_3$ can be represented as $P_3(s) = (\alpha, \beta-2s, \gamma+s)$ for some scalar $s$.

Step 4: Use the intersection condition

Since $L_3$ intersects $L_1$, there must be a point that lies on both lines. Let this point be $P_1(t) = P_3(s)$. Equating the coordinates gives us a system of three equations.

Step 5: Solve for t, s, and the expression

From the system of equations, we can express $\alpha$, $\beta$, and $\gamma$ in terms of $t$ and $s$. We need to find the value of $|5\alpha - 11\beta - 8\gamma|$. Substitute the expressions for $\alpha$, $\beta$, and $\gamma$ into the target expression.

Step 6: Substitute and simplify

Substitute the expressions for $\alpha$, $\beta$, and $\gamma$ into $5\alpha - 11\beta - 8\gamma$. After expanding and combining like terms, the expression simplifies to $-25 - 22t - 14s$.

Step 7: Final value

The problem asks for the absolute value of the expression. Since the problem statement implies a unique value for the expression, and we have simplified it to $-25 - 22t - 14s$, we need to recognize that the problem is designed such that the $t$ and $s$ terms cancel out or are implicitly zero for the specific point of intersection. However, without further constraints on $L_3$ (e.g., it also intersects $L_2$), we cannot determine $t$ and $s$. The problem is likely designed such that the expression evaluates to a constant regardless of $t$ and $s$. Re-checking the calculation, the expression $5\alpha - 11\beta - 8\gamma$ simplifies to $-25 - 22t - 14s$. This indicates that the value depends on $t$ and $s$. This suggests that the problem might be ill-posed or there's a missing piece of information or a common trick. Let's re-evaluate the coefficients. The coefficients of $t$ are $5 - 11 - 16 = -22$. The coefficients of $s$ are $-22 + 8 = -14$. The constant term is $5 - 22 - 8 = -25$. The expression is indeed $-25 - 22t - 14s$. This means the value is not a constant. Let's assume there is a typo in the problem and it should have been a constant. If the problem intended for the expression to be a constant, the coefficients of $t$ and $s$ should have cancelled out. Given the current problem statement, the value is not uniquely determined. However, in competitive exams, such problems usually have a constant answer. Let's assume the question implies that the point $(\alpha, \beta, \gamma)$ is the point of intersection of $L_1$ and $L_3$. The question asks for the value of $|5\alpha - 11\beta - 8\gamma|$. This value is dependent on $t$ and $s$. If the problem implies that $(\alpha, \beta, \gamma)$ is a specific point, for example, the point of intersection of $L_1$ and $L_2$ (if they intersect), or the shortest distance point, then $t$ and $s$ would be fixed. However, $L_1$ and $L_2$ are skew lines. Let's re-read: "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 $L_1$ ". This means $(\alpha, \beta, \gamma)$ is a point on $L_3$. The point of intersection of $L_1$ and $L_3$ is $P_1(t) = (1+t, 2+t, 1+2t)$. The line $L_3$ passes through $(\alpha, \beta, \gamma)$ and has direction vector $(0, -2, 1)$. So, the equation of $L_3$ is $\frac{x-\alpha}{0} = \frac{y-\beta}{-2} = \frac{z-\gamma}{1}$. The point of intersection $(1+t, 2+t, 1+2t)$ must satisfy the equation of $L_3$. This means: $1+t = \alpha$, $\frac{2+t-\beta}{-2} = \frac{1+2t-\gamma}{1}$. From these, we get $2+t-\beta = -2(1+2t-\gamma)$ which is $2+t-\beta = -2-4t+2\gamma$. So, $\beta = 4+5t-2\gamma$. This is not the same as the previous setup. Let's use the fact that $(\alpha, \beta, \gamma)$ is a point on $L_3$. So, $(\alpha, \beta, \gamma)$ is of the form $(x_0, y_0-2s, z_0+s)$ where $(x_0, y_0, z_0)$ is some point on $L_3$. The problem states that $L_3$ passes through $(\alpha, \beta, \gamma)$. This means $(\alpha, \beta, \gamma)$ is a specific point, not a general point on $L_3$. The point of intersection of $L_1$ and $L_3$ is $P_I$. The vector from $(\alpha, \beta, \gamma)$ to $P_I$ must be parallel to $\vec{d}_{3}$. Let $P_I = (1+t, 2+t, 1+2t)$. Then $\vec{P_I - (\alpha, \beta, \gamma)} = (1+t-\alpha, 2+t-\beta, 1+2t-\gamma)$ must be parallel to $(0, -2, 1)$. This means: $1+t-\alpha = 0 \implies \alpha = 1+t$. $\frac{2+t-\beta}{-2} = \frac{1+2t-\gamma}{1} = k$ (for some scalar $k$). So, $2+t-\beta = -2k$ and $1+2t-\gamma = k$. This implies $2+t-\beta = -2(1+2t-\gamma)$. $2+t-\beta = -2-4t+2\gamma$. $\beta = 4+5t-2\gamma$. Now substitute these into $5\alpha - 11\beta - 8\gamma$: $5(1+t) - 11(4+5t-2\gamma) - 8\gamma = 5+5t - 44-55t+22\gamma - 8\gamma = -39 - 50t + 14\gamma$. This still depends on $t$ and $\gamma$. This indicates that the problem is underdetermined or there's a standard interpretation I'm missing. Let's consider the case where $(\alpha, \beta, \gamma)$ is the foot of the perpendicular from a point on $L_2$ to $L_1$. No, it says $L_3$ passes through $(\alpha, \beta, \gamma)$. The most common interpretation for such problems is that $(\alpha, \beta, \gamma)$ is the point of intersection of $L_1$ and $L_3$. If $(\alpha, \beta, \gamma)$ is the point of intersection of $L_1$ and $L_3$, then $(\alpha, \beta, \gamma) = (1+t, 2+t, 1+2t)$ for some $t$. In this case, $L_3$ passes through $(\alpha, \beta, \gamma)$ and has direction vector $(0, -2, 1)$. This means the point $(\alpha, \beta, \gamma)$ is on $L_1$ and $L_3$. So, $\alpha = 1+t$, $\beta = 2+t$, $\gamma = 1+2t$. Now substitute these into the expression: $5(1+t) - 11(2+t) - 8(1+2t) = 5+5t - 22-11t - 8-16t = (5-22-8) + (5-11-16)t = -25 - 22t$. This still depends on $t$. This is a common type of problem where the expression simplifies to a constant. Let's re-check the cross product. $\vec{d}_{3} = (0, -2, 1)$. This is correct. Let's re-check the problem statement. " $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$ ". This means the point of intersection of $L_3$ and $L_1$ is $P_I = (1+t, 2+t, 1+2t)$. The line $L_3$ passes through $(\alpha, \beta, \gamma)$ and $P_I$. So the vector $\vec{(\alpha, \beta, \gamma)P_I}$ is parallel to $\vec{d}_{3}$. Thus, $(1+t-\alpha, 2+t-\beta, 1+2t-\gamma) = k(0, -2, 1)$ for some scalar $k$. This implies: $1+t-\alpha = 0 \implies \alpha = 1+t$. $2+t-\beta = -2k$. $1+2t-\gamma = k$. From the last two equations, $2+t-\beta = -2(1+2t-\gamma)$. $2+t-\beta = -2-4t+2\gamma$. $\beta = 4+5t-2\gamma$. Now substitute $\alpha = 1+t$ and $\beta = 4+5t-2\gamma$ into the expression $5\alpha - 11\beta - 8\gamma$: $5(1+t) - 11(4+5t-2\gamma) - 8\gamma = 5+5t - 44-55t+22\gamma - 8\gamma = -39 - 50t + 14\gamma$. This still depends on $t$ and $\gamma$. This problem is underdetermined as stated. However, if the question implies that $(\alpha, \beta, \gamma)$ is the foot of the perpendicular from a point on $L_2$ to $L_1$, then it would be a specific point. But the wording is " $L_3$ passes through $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$ ". This means $(\alpha, \beta, \gamma)$ is a point on $L_3$. Let's consider the case where $(\alpha, \beta, \gamma)$ is the point of intersection of $L_1$ and $L_3$. Then $(\alpha, \beta, \gamma) = (1+t, 2+t, 1+2t)$ for some $t$. The line $L_3$ passes through this point and has direction vector $(0, -2, 1)$. This is consistent. So, $\alpha = 1+t$, $\beta = 2+t$, $\gamma = 1+2t$. The expression becomes $5(1+t) - 11(2+t) - 8(1+2t) = 5+5t - 22-11t - 8-16t = -25 - 22t$. This is not a constant. There must be a misunderstanding of the problem statement or a typo in the problem. Let's assume the problem meant that $(\alpha, \beta, \gamma)$ is the point of intersection of $L_1$ and $L_3$, and that $L_3$ is the shortest distance line between $L_1$ and $L_2$. In that case, $L_3$ would intersect $L_1$ at $(\alpha, \beta, \gamma)$ and $L_2$ at some other point. Let $P_1(t) = (1+t, 2+t, 1+2t)$ and $P_2(u) = (-1+u, 2+2u, 4u)$. The vector $\vec{P_1P_2} = (-2+u-t, 2u-t, 4u-1-2t)$ must be parallel to $\vec{d}_{3} = (0, -2, 1)$. So, $-2+u-t = 0 \implies u-t=2$. $\frac{2u-t}{-2} = \frac{4u-1-2t}{1}$. $2u-t = -2(4u-1-2t) = -8u+2+4t$. $10u - 5t = 2$. We have a system of equations: $u-t=2$ and $10u-5t=2$. From the first equation, $u=t+2$. Substitute into the second: $10(t+2)-5t=2 \implies 10t+20-5t=2 \implies 5t = -18 \implies t = -18/5$. Then $u = -18/5 + 2 = -8/5$. So, the point of intersection on $L_1$ is $(\alpha, \beta, \gamma) = (1-18/5, 2-18/5, 1+2(-18/5)) = (-13/5, -8/5, -31/5)$. Now substitute these values into $5\alpha - 11\beta - 8\gamma$: $5(-13/5) - 11(-8/5) - 8(-31/5) = -13 + 88/5 + 248/5 = -13 + 336/5 = (-65+336)/5 = 271/5$. The absolute value is $271/5$. This is a constant. This interpretation makes the problem solvable. The problem statement is slightly ambiguous, but this is the standard interpretation for such problems in competitive exams. The line $L_3$ is the common perpendicular between $L_1$ and $L_2$, and $(\alpha, \beta, \gamma)$ is the foot of this perpendicular on $L_1$.

Step 8: Calculate the final value

Substitute the values of $\alpha = -13/5$, $\beta = -8/5$, and $\gamma = -31/5$ into the expression $5\alpha - 11\beta - 8\gamma$. Simplify the expression to get $271/5$. The absolute value of $271/5$ is $271/5$.

More from this chapter

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• Find the equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes x + 2y + 3z = 4 and 2x - 3y + 4z = 5.
• Let the line passing through the points (-1, 2, 1) and parallel to the line (x-1)/(2) = (y+1)/(3) = (z)/(4) intersect the line (x+2)/(3) = (y-3)/(2) = (z-4)/(3)
• Let the shortest distance between the lines (x-3)/(3) = (y-α)/(-1) = (z-3)/(1) and (x+3)/(-3) = (y+7)/(2) = (z-β)/(4) be 3√30. Then the positive value of 5α + β
• Consider the lines L_1: x - 1 = y - 2 = z and L_2: x - 2 = y - 2 = z - 1. Let the feet of the perpendiculars from the point P(5,1,-3) on the lines L_1 and L_2 b

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