Question

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line \[\frac{x - 0}{1} = \frac{y - 3}{1} = \frac{z - 1}{-1}\]and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$, then $K$ is equal to:

• A. 36
• B. 72
• C. 18
• D. 81

Answer 1

The Correct Option is D

The objective is to first determine the coordinates of point \(P(\alpha, \beta, \gamma)\), which is the reflection of point \(Q(3, -3, 1)\) across the provided line. Subsequently, we must compute the area, denoted as \(\lambda\), of the triangle formed by points \(P\), \(Q\), and \(R(2, 5, -1)\). Finally, employing the given equation \(\lambda^2 = 14K\), we are to ascertain the value of \(K\).

Concept Used:

The resolution of this problem relies on fundamental concepts from 3D geometry:

1. Image of a point with respect to a line: To find the image \(P\) of a point \(Q\) relative to a line, we first locate the foot of the perpendicular, \(M\), from \(Q\) to the line. Point \(M\) serves as the midpoint of the line segment \(PQ\).
2. Foot of the perpendicular: A general point \(M\) on the line is parameterized. The direction ratios of the line segment \(QM\) are derived. Given that \(QM\) is orthogonal to the specified line, the dot product of their direction ratios is zero, enabling the determination of the parameter and subsequently the coordinates of \(M\).
3. Midpoint formula: For a midpoint \(M(x_m, y_m, z_m)\) of a segment connecting \(P(\alpha, \beta, \gamma)\) and \(Q(x_q, y_q, z_q)\), the coordinates satisfy \(x_m = (\alpha + x_q)/2\), \(y_m = (\beta + y_q)/2\), and \(z_m = (\gamma + z_q)/2\).
4. Area of a triangle in 3D: The area of a triangle with vertices \(P\), \(Q\), and \(R\) is calculated as half the magnitude of the cross product of two vectors representing adjacent sides, for instance, \(\frac{1}{2} |\vec{QP} \times \vec{QR}|\).

Step-by-Step Solution:

Step 1: Determine the foot of the perpendicular from point \(Q\) to the line.

The line \(L\) is defined by \( \frac{x}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t\). A generic point on this line is \(M(t, 3+t, 1-t)\). Let \(M\) be the foot of the perpendicular from \(Q(3, -3, 1)\) to \(L\).

The direction ratios of the line segment \(QM\) are:

\[\langle (t-3), (3+t - (-3)), (1-t - 1) \rangle = \langle t-3, t+6, -t \rangle\]

The direction ratios of line \(L\) are \(\langle 1, 1, -1 \rangle\).

Since \(QM \perp L\), their direction ratios' dot product is zero:

\[(1)(t-3) + (1)(t+6) + (-1)(-t) = 0\]\[t - 3 + t + 6 + t = 0\]\[3t + 3 = 0 \implies t = -1\]

Substituting \(t = -1\) into the coordinates of \(M\) yields the foot of the perpendicular:

\[M = (-1, 3-1, 1-(-1)) = (-1, 2, 2)\]

Step 2: Ascertain the coordinates of the image \(P(\alpha, \beta, \gamma)\).

Point \(M\) is the midpoint of segment \(PQ\). Applying the midpoint formula:

\[M\left(\frac{\alpha+3}{2}, \frac{\beta-3}{2}, \frac{\gamma+1}{2}\right) = (-1, 2, 2)\]

Equating corresponding coordinates:

\[\frac{\alpha+3}{2} = -1 \implies \alpha+3 = -2 \implies \alpha = -5\]\[\frac{\beta-3}{2} = 2 \implies \beta-3 = 4 \implies \beta = 7\]\[\frac{\gamma+1}{2} = 2 \implies \gamma+1 = 4 \implies \gamma = 3\]

Thus, the image point is \(P(-5, 7, 3)\).

Step 3: Compute the area of triangle \(PQR\).

The vertices are \(P(-5, 7, 3)\), \(Q(3, -3, 1)\), and \(R(2, 5, -1)\). The vectors \(\vec{QP}\) and \(\vec{QR}\) are:

\[\vec{QP} = P - Q = \langle -5-3, 7-(-3), 3-1 \rangle = \langle -8, 10, 2 \rangle\]\[\vec{QR} = R - Q = \langle 2-3, 5-(-3), -1-1 \rangle = \langle -1, 8, -2 \rangle\]

The area of \(\triangle PQR\) is \(\lambda = \frac{1}{2} |\vec{QP} \times \vec{QR}|\).

Step 4: Calculate the cross product \(\vec{QP} \times \vec{QR}\).

\[\vec{QP} \times \vec{QR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 10 & 2 \\ -1 & 8 & -2 \end{vmatrix}\]\[= \mathbf{i}((10)(-2) - (2)(8)) - \mathbf{j}((-8)(-2) - (2)(-1)) + \mathbf{k}((-8)(8) - (10)(-1))\]\[= \mathbf{i}(-20 - 16) - \mathbf{j}(16 + 2) + \mathbf{k}(-64 + 10)\]\[= -36\mathbf{i} - 18\mathbf{j} - 54\mathbf{k}\]

Step 5: Determine the magnitude of the cross product and the area \(\lambda\).

\[|\vec{QP} \times \vec{QR}| = \sqrt{(-36)^2 + (-18)^2 + (-54)^2}\]\[= \sqrt{1296 + 324 + 2916} = \sqrt{4536}\]

Simplifying by factoring out \(18^2 = 324\):

\[\sqrt{324 \cdot 4 + 324 \cdot 1 + 324 \cdot 9} = \sqrt{324(4+1+9)} = \sqrt{324 \cdot 14} = 18\sqrt{14}\]

The area is:

\[\lambda = \frac{1}{2} |\vec{QP} \times \vec{QR}| = \frac{1}{2} (18\sqrt{14}) = 9\sqrt{14}\]

Step 6: Calculate the value of \(K\).

Given the relation \(\lambda^2 = 14K\), substitute the value of \(\lambda\):

\[(9\sqrt{14})^2 = 14K\]\[81 \times 14 = 14K\]\[K = 81\]

Consequently, the value of \(K\) is 81.