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 problem requires us to first find the coordinates of point \(P(\alpha, \beta, \gamma)\), which is the image of point \(Q(3, -3, 1)\) with respect to the given line. Then, we need to calculate the area, \(\lambda\), of the triangle formed by points \(P\), \(Q\), and \(R(2, 5, -1)\). Finally, using the given relation \(\lambda^2 = 14K\), we must find the value of \(K\).

Concept Used:

The solution involves the following concepts from 3D geometry:

1. Image of a point in a line: To find the image \(P\) of a point \(Q\) in a line, we first find the foot of the perpendicular, say \(M\), from \(Q\) to the line. The point \(M\) is the midpoint of the line segment \(PQ\).
2. Foot of the perpendicular: A general point \(M\) on the line is taken using a parameter. The direction ratios of the line segment \(QM\) are found. Since \(QM\) is perpendicular to the given line, the dot product of their direction ratios is zero. This allows us to find the parameter and hence the coordinates of \(M\).
3. Midpoint formula: If \(M(x_m, y_m, z_m)\) is the midpoint of \(P(\alpha, \beta, \gamma)\) and \(Q(x_q, y_q, z_q)\), then \(x_m = (\alpha + x_q)/2\), \(y_m = (\beta + y_q)/2\), 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 given by half the magnitude of the cross product of two vectors forming adjacent sides of the triangle, e.g., \(\frac{1}{2} |\vec{QP} \times \vec{QR}|\).

Step-by-Step Solution:

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

The given line is \(L: \frac{x}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t\). Any point on this line can be represented as \(M(t, 3+t, 1-t)\). Let \(M\) be the foot of the perpendicular from \(Q(3, -3, 1)\) to the line \(L\).

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 the line \(L\) are \(\langle 1, 1, -1 \rangle\).

Since \(QM\) is perpendicular to \(L\), the dot product of their direction ratios 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 \(M\), we get the foot of the perpendicular:

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

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

The point \(M\) is the midpoint of the segment \(PQ\). Using the midpoint formula:

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

Equating the coordinates:

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

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

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

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

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

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

Step 4: Compute 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: Find the magnitude of the cross product and the area \(\lambda\).

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

We can simplify this 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 \(K\).

We are given the relation \(\lambda^2 = 14K\). Substituting the value of \(\lambda\):

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

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

Answer 2

The coordinates of \( Q \) are \( (3, -3, 1) \) and \( R \) is at \( (2, 5, -1) \).

Step 1: Calculating \( RQ \):

\[ RQ = \sqrt{(2 - 3)^2 + (5 + 3)^2 + (-1 - 1)^2} = \sqrt{1 + 64 + 4} = \sqrt{69} \]

Step 2: Representing \( \vec{RQ} \):

\[ \vec{RQ} = -\hat{i} + 8\hat{j} - 2\hat{k} \]

Step 3: Representing \( \vec{RS} \):

\[ \vec{RS} = \hat{i} + \hat{j} - \hat{k} \]

Step 4: Finding cosine of the angle \( \theta \) between vectors \( \vec{RQ} \) and \( \vec{RS} \):

\[ \cos \theta = \frac{\vec{RQ} \times \vec{RS}}{|\vec{RQ}||\vec{RS}|} \]

\[ = \frac{(-1 \times 1) + (8 \times 1) + (-2 \times -1)}{\sqrt{69} \times \sqrt{3}} = \frac{-1 + 8 + 2}{\sqrt{69} \times \sqrt{3}} = \frac{9}{3\sqrt{23}} \]

Step 5: Using sine of the angle:

\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \frac{\sqrt{23}}{\sqrt{69}} \]

Step 6: Finding area of triangle \( PQR \):

\[ \text{Area} = \frac{1}{2} \times |\vec{RQ} \times \vec{RS}| \times \sin \theta = \frac{1}{2} \times \sqrt{69} \times \sqrt{3} \times \frac{\sqrt{23}}{\sqrt{69}} = \frac{\sqrt{3} \times \sqrt{23}}{2} \]

Step 7: Given \( \lambda^2 = 14K \):

\[ \lambda^2 = 81.14 = 14K \implies K = 81 \]