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

Answer: (D)

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 $\lambda^2 = 14K$:

$\lambda^2 = 81.14 = 14K \implies K = 81$