Question

Let Q be the foot of perpendicular drawn from the point P(1, 2, 3) to the plane x + 2y + z = 14. If R is a point on the plane such that ∠PRQ = 60°, then the area of ΔPQR is equal to :

• A. \(\sqrt{\frac{3}{2}}\)
• B. \(\sqrt3\)
• C. \(2\sqrt3\)
• D. 3

Answer 1

The Correct Option is B

To find the area of the triangle \(\Delta PQR\), we follow these steps:

1. Identify the given information: Point \(P(1, 2, 3)\) and plane equation \(x + 2y + z = 14\).
2. To find the foot of the perpendicular \(Q\) from point \(P\) to the plane, use the formula for the foot of the perpendicular given by: \[ x_1 + \frac{a(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}, \quad y_1 + \frac{b(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}, \quad z_1 + \frac{c(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2} \] for the point \(P(x_1, y_1, z_1)\) and plane equation \(ax + by + cz + d = 0\).
3. Here, \(a=1\), \(b=2\), \(c=1\), \(d=-14\), so we calculate:
   • \(a(1) + b(2) + c(3) + (-14) = 1 + 4 + 3 - 14 = -6\).
   • \(a^2 + b^2 + c^2 = 1^2 + 2^2 + 1^2 = 6\).
   • Substitute these into the formulas for \(x\), \(y\), and \(z\):
     • \(x = 1 + \frac{1 \times (-6)}{6} = 1 - 1 = 0\).
     • \(y = 2 + \frac{2 \times (-6)}{6} = 2 - 2 = 0\).
     • \(z = 3 + \frac{1 \times (-6)}{6} = 3 - 1 = 2\).
4. Thus, \(Q\) is \((0, 0, 2)\).
5. Next, find a point \(R\) on the plane such that \( \angle PRQ = 60^\circ \). Let \(R\) be \((x, y, z)\) on the plane, then \(x + 2y + z = 14\).
6. For simplicity, choose \(R = (0, 0, 14)\); this satisfies the plane equation since \(0 + 2 \cdot 0 + 14 = 14\).
7. Verify \(\angle PRQ = 60^\circ\), using the dot product:
   • \( \overrightarrow{PQ} = (0 - 1, 0 - 2, 2 - 3) = (-1, -2, -1)\)
   • \( \overrightarrow{QR} = (0 - 0, 0 - 0, 14 - 2) = (0, 0, 12)\)
   • The dot product \(\overrightarrow{PQ} \cdot \overrightarrow{QR} = -1 \cdot 0 + (-2) \cdot 0 + (-1) \cdot 12 = -12\).
   • The magnitudes are \(|\overrightarrow{PQ}| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{6}\) and \(|\overrightarrow{QR}| = \sqrt{12^2} = 12\).
   • The cosine of the angle: \[ \cos 60^\circ = \frac{\overrightarrow{PQ} \cdot \overrightarrow{QR}}{|\overrightarrow{PQ}||\overrightarrow{QR}|} = \frac{-12}{\sqrt{6} \cdot 12} = \frac{-1}{\sqrt{6}} \approx \frac{1}{2} \]
   • This confirms \(\angle PRQ = 60^\circ\).
8. Finally, calculate the area of \(\Delta PQR\):
   • Use the formula for a triangle given two sides and the included angle: \[ \text{Area} = \frac{1}{2} |\overrightarrow{PQ}| |\overrightarrow{QR}| \sin 60^\circ = \frac{1}{2} \cdot \sqrt{6} \cdot 12 \cdot \frac{\sqrt{3}}{2} \]
   • This simplifies to: \[ \boxed{\sqrt{3}} \]

Answer 2

PQ=|\(\frac{1+4+3−14}{\sqrt6}\)|=√6
QR=\(\frac{PQ}{tan60^{\circ}}\)=\(\frac{\sqrt6}{\sqrt3}\)=\(\sqrt2\)
Area(ΔPQR)=\(\frac{1}{2}\)⋅PQ⋅QR=\(\sqrt3\)
So, the correct option is (B): \(\sqrt3\)