Question

Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$ Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$ If $\alpha$ is the area of triangle $P Q R$, then $\alpha^2$ is equal to

Hint:

To find the area of a triangle given by three points in space, use the formula \( \frac{1}{2} \left| \mathbf{QR} \times \mathbf{QP} \right| \), where \( \mathbf{QR} \) and \( \mathbf{QP} \) are the position vectors of the sides of the triangle.

Answer 1

Correct Answer: 180

To solve the problem, we need to find the area of the triangle formed by points \(P\), \(Q\), and \(R\) and then compute \(\alpha^2\), where \(\alpha\) is the area. Here's a step-by-step approach:

Step 1: Find the intersection point \(P\) of line \(L\) and plane

The line \(L\) is given parametrically as \(x=1+2t\), \(y=-1-t\), \(z=3+t\). Substituting these in the plane equation \(2x+y+3z=16\), we have:

\(2(1+2t) + (-1-t) + 3(3+t) = 16\)

Solving,
\(2 + 4t - 1 - t + 9 + 3t = 16\)
\(10 + 6t = 16\)
\(6t = 6\)
\(t = 1\)

Thus, point \(P\) is:
\(x = 1+2(1)=3\)
\(y = -1-1=-2\)
\(z = 3+1=4\)
\(P(3,-2,4)\)

Step 2: Find the foot of perpendicular \(Q\) from \(R\) to line \(L\)

Point on the line \(L\) for parametric \(\lambda\) is \((1+2\lambda, -1-\lambda, 3+\lambda)\).

The vector \(RQ\) is orthogonal to \(L\). The directional vector of \(L\) is \((2, -1, 1)\). \(RQ\) equals \((2\lambda, -\lambda, 6+\lambda)\), derived from \(\overrightarrow{RQ} = (1+2\lambda-1, -1-\lambda+1, 3+\lambda+3)\).

Dot product of \(\overrightarrow{RQ}\) and line direction vector must be zero:
\(2(2\lambda) + (-1)(-\lambda) + 1(6+\lambda) = 0\)
\(4\lambda + \lambda + 6 + \lambda = 0\)
\(6\lambda + 6 = 0\)
\(\lambda = -1\)

Substitute \(\lambda = -1\) to find \(Q\):
\(x = 1+2(-1)=-1\),
\(y = -1+1=0\),
\(z = 3-1=2\)
\(Q(-1,0,2)\)

Step 3: Calculate area of \(\triangle PQR\) and \(\alpha^2\)

Coordinates: \(P(3,-2,4)\), \(Q(-1,0,2)\), \(R(1,-1,-3)\).

Use the formula for the area of a triangle given vertices \((x_1,y_1,z_1)\), \((x_2,y_2,z_2)\), \((x_3,y_3,z_3)\):
\(\text{Area} = \frac{1}{2} \sqrt{\left|\begin{array}{ccc} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right|^2}\)

Setup vectors: \(\overrightarrow{PQ} = (-4,2,-2)\), \(\overrightarrow{PR} = (-2,1,-7)\)

Cross product \(\overrightarrow{PQ} \times \overrightarrow{PR} =\)
\(i((2)(-7) - (1)(-2)) - j((-4)(-7) - (-2)(-2)) + k((-4)(1) - 2(-2))\)
\(= i(-14 + 2) - j(28 - 4) + k(-4 + 4)\)
\(= (-12, -24, 0)\)

Magnitude: \(|(-12, -24, 0)| = \sqrt{144 + 576} = \sqrt{720} = 12\sqrt{5}\)

Area \(\triangle PQR = \frac{1}{2}(12\sqrt{5}) = 6\sqrt{5}\)

\(\alpha = 6\sqrt{5}\), \(\alpha^2 = (6\sqrt{5})^2 = 36 \times 5 = 180\)

Thus, the solution is \(180\), which fits the given range.