Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $P(x_1, y_1, z_1)$ with a direction vector $\vec{b} = (a, b, c)$ can be represented parametrically as $x = x_1 + a\lambda$, $y = y_1 + b\lambda$, $z = z_1 + c\lambda$, where $\lambda$ is a scalar parameter.
• Direction Vector of Line of Intersection of Two Planes: If a line is the intersection of two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$, its direction vector is given by the cross product $\vec{n_1} \times \vec{n_2}$.
• Foot of Perpendicular and Distance from a Point to a Line: To find the foot of the perpendicular $N$ from an external point $Q$ to a line $L$, we first express a general point on $L$ parametrically. The vector $\vec{QN}$ connecting $Q$ to $N$ must be perpendicular to the direction vector of line $L$. This condition is expressed by their dot product being zero: $\vec{QN} \cdot \vec{b} = 0$. Once $N$ is found, the length of the perpendicular is the distance between $Q$ and $N$ using the 3D distance formula: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Step-by-Step Solution

We need to find the length of the perpendicular from the point $Q(1,-2,5)$ to a line $L$. The line $L$ passes through $P(1,2,4)$ and is parallel to the line of intersection of the planes $x+y-z=0$ and $x-2y+3z-5=0$.

Step 1: Determine the Direction Vector of Line $L$

• What we are doing: We need the direction vector of line $L$ to write its equation. Since line $L$ is parallel to the line of intersection of two planes, its direction vector will be the same as the direction vector of that intersecting line.

• Why we are doing it: The direction vector is a fundamental component for defining a line in 3D space.

• Explanation: The direction vector of the line of intersection of two planes is perpendicular to the normal vectors of both planes. Thus, it can be found by taking the cross product of their normal vectors.

  • From Plane 1: $x+y-z=0$, the normal vector is $\vec{n_1} = (1,1,-1)$.
  • From Plane 2: $x-2y+3z-5=0$, the normal vector is $\vec{n_2} = (1,-2,3)$.

  Now, we calculate the cross product $\vec{n_1} \times \vec{n_2}$ to find the direction vector $\vec{b}$ of line $L$: $\vec{b} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 1 & -1 \\ 1 & -2 & 3 \end{vmatrix}$ Expanding the determinant: $\vec{b} = \widehat i ((1)(3) - (-1)(-2)) - \widehat j ((1)(3) - (-1)(1)) + \widehat k ((1)(-2) - (1)(1))$ $\vec{b} = \widehat i (3 - 2) - \widehat j (3 + 1) + \widehat k (-2 - 1)$ $\vec{b} = 1\widehat i - 4\widehat j - 3\widehat k$ Thus, the direction vector of the line $L$ is $\vec{b} = (1, -4, -3)$.

Step 2: Formulate the Parametric Equation of Line $L$

• What we are doing: We are writing the equation of line $L$ in parametric form.
• Why we are doing it: A parametric equation allows us to represent any general point on the line using a single parameter, which is essential for the next step of finding the foot of the perpendicular.
• Explanation: Line $L$ passes through point $P(1,2,4)$ and has the direction vector $\vec{b} = (1, -4, -3)$ (from Step 1). Using the parametric form $x = x_1 + a\lambda$, $y = y_1 + b\lambda$, $z = z_1 + c\lambda$: Any general point $N$ on line $L$ can be expressed as: $N(\lambda) = (1 + 1\lambda, 2 - 4\lambda, 4 - 3\lambda)$ $N(\lambda) = (1+\lambda, 2-4\lambda, 4-3\lambda)$

Step 3: Find the Foot of the Perpendicular

• What we are doing: We are finding the specific point on line $L$ that is closest to point $Q(1,-2,5)$. This point is called the foot of the perpendicular.
• Why we are doing it: The distance from $Q$ to line $L$ is the distance from $Q$ to this foot of the perpendicular.
• Explanation: Let $Q(1,-2,5)$ be the given external point and $N(\lambda) = (1+\lambda, 2-4\lambda, 4-3\lambda)$ be a general point on line $L$. The vector $\vec{QN}$ connects point $Q$ to point $N$. $\vec{QN} = N - Q$ $\vec{QN} = ((1+\lambda) - 1, (2-4\lambda) - (-2), (4-3\lambda) - 5)$ $\vec{QN} = (\lambda, 2-4\lambda+2, 4-3\lambda-5)$ $\vec{QN} = (\lambda, 4-4\lambda, -1-3\lambda)$ For $N$ to be the foot of the perpendicular, the vector $\vec{QN}$ must be perpendicular to the direction vector of line $L$, which is $\vec{b} = (1, -4, -3)$. Their dot product must be zero: $\vec{QN} \cdot \vec{b} = 0$. $(\lambda)(1) + (4-4\lambda)(-4) + (-1-3\lambda)(-3) = 0$ $\lambda - 16 + 16\lambda + 3 + 9\lambda = 0$ Combine like terms: $(1 + 16 + 9)\lambda + (-16 + 3) = 0$ $26\lambda - 13 = 0$ $26\lambda = 13$ $\lambda = \frac{13}{26} = \frac{1}{2}$ Now, substitute $\lambda = \frac{1}{2}$ back into the coordinates of $N(\lambda)$ to find the exact coordinates of the foot of the perpendicular: $N = (1+\frac{1}{2}, 2-4(\frac{1}{2}), 4-3(\frac{1}{2}))$ $N = (\frac{3}{2}, 2-2, 4-\frac{3}{2})$ $N = (\frac{3}{2}, 0, \frac{8-3}{2})$ $N = (\frac{3}{2}, 0, \frac{5}{2})$ So, the foot of the perpendicular is $N(\frac{3}{2}, 0, \frac{5}{2})$.

Step 4: Calculate the Length of the Perpendicular

• What we are doing: We are calculating the distance between point $Q$ and the foot of the perpendicular $N$.
• Why we are doing it: This distance is the required length of the perpendicular.
• Explanation: The length of the perpendicular is the distance between $Q(1,-2,5)$ and $N(\frac{3}{2}, 0, \frac{5}{2})$. Using the 3D distance formula: $QN = \sqrt{\left(\frac{3}{2}-1\right)^2 + (0-(-2))^2 + \left(\frac{5}{2}-5\right)^2}$ $QN = \sqrt{\left(\frac{3-2}{2}\right)^2 + (2)^2 + \left(\frac{5-10}{2}\right)^2}$ $QN = \sqrt{\left(\frac{1}{2}\right)^2 + (2)^2 + \left(-\frac{5}{2}\right)^2}$ $QN = \sqrt{\frac{1}{4} + 4 + \frac{25}{4}}$ To sum these terms, find a common denominator: $QN = \sqrt{\frac{1}{4} + \frac{16}{4} + \frac{25}{4}}$ $QN = \sqrt{\frac{1+16+25}{4}}$ $QN = \sqrt{\frac{42}{4}}$ $QN = \sqrt{\frac{21}{2}}$

Common Mistakes & Tips

• Cross Product Calculation: Be meticulous with signs and order of operations when computing the determinant for the cross product. A common error is forgetting the negative sign for the $\widehat{j}$ component's minor.
• Algebraic Errors with Parameter $\lambda$: Ensure careful distribution and combining of like terms when solving the dot product equation for $\lambda$.
• Fraction Arithmetic: Double-check calculations involving fractions, especially when squaring and summing them in the distance formula. Convert all terms to a common denominator before summing.
• Understanding Perpendicularity: Remember that the vector from the external point to the foot of the perpendicular is perpendicular to the line's direction vector, not necessarily to the line itself in general.

Summary

To find the length of the perpendicular from a point to a line, we first determined the direction vector of the line by taking the cross product of the normal vectors of the planes it's parallel to. We then formulated the parametric equation of the line using the given point it passes through. Next, we found the foot of the perpendicular by setting the dot product of the vector from the external point to a general point on the line and the line's direction vector to zero. Finally, we calculated the distance between the external point and this foot of the perpendicular using the 3D distance formula. The calculated length is $\sqrt{\frac{21}{2}}$.

The final answer is $\boxed{\sqrt{\frac{21}{2}}}$, which corresponds to option (A).