<p>The vertices \(B\) and \(C\) of a \(\triangle ABC\) lie on the line, \(\dfrac{x+2}{3} = \dfrac{y-1}{0} = \dfrac{z}{4}\) such that \(BC = 5\) units. Then the area (in sq. units) of this triangle, given that the point \(A(1, -1, 2)\), is ______ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: The area of triangle ABC equals (1/2)|AB × AC|, where B and C lie on a given line. Since B and C are collinear on the line with direction vector (3,0,4), we can parameterize them and use the constraint |BC| = 5 to find their positions, then compute the cross product.
Step 1: Find the perpendicular distance from A(1, -1, 2) to the line. The line is: r = (-2, 1, 0) + t(3, 0, 4). Direction vector d = (3, 0, 4), | d | = 5. Step 2: Vector from point on line to A: Take point P_0(-2, 1, 0) on the line. PA = (1-(-2), -1-1, 2-0) = (3, -2, 2) Step 3: Perpendicular distance = | PA × d | / | d | PA × d = (3, -2, 2) × (3, 0, 4) = ((-2)(4) - (2)(0), (2)(3) - (3)(4), (3)(0) - (-2)(3)) = (-8, 6-12, 0+6) = (-8, -6, 6) | PA × d | = √(64 + 36 + 36) = √136 = 2√34 Step 4: Perpendicular distance h = 2√34 / 5 Step 5: Area of triangle = (1/2) × base × height = (1/2) × BC × h = (1/2) × 5 × (2√34/5) = √34 ∴ Answer: 5.831 sq. units
Correct Answer: 5