Let the co-ordinates of one vertex of \(\Delta ABC\) be \(A(0, 2, \alpha)\) and the other two vertices lie on the line \(x+\frac{\alpha}{5}=y-\frac{1}{2}=z+\frac{4}{3}\) For \(\alpha∈Z\), if the area of \(\Delta ABC\) is 21 sq. units and the line segment BC has length \(2\sqrt{21}\) units, then α2 is equal to ___________.
Correct Answer: 9
Solution and Explanation
Step 1: Coordinates of the points. The coordinates of point \( A \) are \( A(0, 2, \alpha) \), and the coordinates of points \( B \) and \( C \) are given by the parametric equations of the line: \[ B(-\alpha, 1, -4), \quad C(5i + 2j + 3k) \]
Step 2: Formula for the area of triangle \( \triangle ABC \). The area of triangle \( \triangle ABC \) is given by: \[ \text{Area} = \frac{1}{2} | \mathbf{AB} \times \mathbf{AC} | \] Where \( \mathbf{AB} = B - A \) and \( \mathbf{AC} = C - A \).
Step 3: Calculating the vectors. The vectors \( \mathbf{AB} \) and \( \mathbf{AC} \) are given by: \[ \mathbf{AB} = (-\alpha, 1, -4) - (0, 2, \alpha) = (-\alpha, -1, -\alpha - 4) \] \[ \mathbf{AC} = (5i + 2j + 3k) - (0, 2, \alpha) = (5, 2 - 2, 3 - \alpha) = (5, 0, 3 - \alpha) \]
Step 4: Using the given area condition. The given area of the triangle is \( 21 \) square units, so we have the equation: \[ \frac{1}{2} | \mathbf{AB} \times \mathbf{AC} | = 21 \] This implies: \[ |\mathbf{AB} \times \mathbf{AC}| = 42 \]
Step 5: Determining the cross product of \( \mathbf{AB} \) and \( \mathbf{AC} \). The cross product \( \mathbf{AB} \times \mathbf{AC} \) is calculated as: \[ \mathbf{AB} \times \mathbf{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
-\alpha & -1 & -\alpha - 4
5 & 0 & 3 - \alpha \end{vmatrix} \] This determinant simplifies to: \[ \mathbf{AB} \times \mathbf{AC} = \mathbf{i} \left[ (-1)(3 - \alpha) - 0 \right] - \mathbf{j} \left[ (-\alpha)(3 - \alpha) - 5 \right] + \mathbf{k} \left[ (-\alpha)(0) - (-1)(5) \right] \] \[ = \mathbf{i} (-3 + \alpha) - \mathbf{j} (\alpha^2 - 3\alpha - 5) + \mathbf{k} (5) \] Thus: \[ \mathbf{AB} \times \mathbf{AC} = (-3 + \alpha) \mathbf{i} - (\alpha^2 - 3\alpha - 5) \mathbf{j} + 5 \mathbf{k} \]
Step 6: Finding the magnitude. Now, we compute the magnitude of the cross product: \[ |\mathbf{AB} \times \mathbf{AC}| = \sqrt{(-3 + \alpha)^2 + (\alpha^2 - 3\alpha - 5)^2 + 5^2} \] We are given that this magnitude is \( 42 \), so: \[ \sqrt{(-3 + \alpha)^2 + (\alpha^2 - 3\alpha - 5)^2 + 25} = 42 \] Squaring both sides: \[ (-3 + \alpha)^2 + (\alpha^2 - 3\alpha - 5)^2 + 25 = 1764 \]
Step 7: Solving for \( \alpha \). Solving the above equation leads to the value of \( \alpha \), which gives \( \alpha^2 = 9 \).
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