Question

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 ___________.

Answer 1

Correct Answer: 9

 To solve the problem, let's break it down into key steps:

1. **Understand the line equation**: The line is given as \(x+\frac{\alpha}{5}=y-\frac{1}{2}=z+\frac{4}{3}=t\). Expressing \(B\) and \(C\) coordinates in terms of \(t\), we have:

\(B(t_1)\): \((t_1-\frac{\alpha}{5},\frac{1}{2}+t_1,-\frac{4}{3}+t_1)\)

\(C(t_2)\): \((t_2-\frac{\alpha}{5},\frac{1}{2}+t_2,-\frac{4}{3}+t_2)\)

2. **Calculate BC length**: Given \(BC=2\sqrt{21}\), use the distance formula:

\[\sqrt{(t_2-t_1)^2+(t_2-t_1)^2+(t_2-t_1)^2}=2\sqrt{21}\]

This simplifies to \(t_2-t_1=2\).

3. **Area of triangle**: Use the formula for the area of \(\Delta ABC\) with vectors:

Vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) are:

\(\overrightarrow{AB}=((t_1-\frac{\alpha}{5}),(\frac{1}{2}+t_1)-2,(-\frac{4}{3}+t_1)-\alpha)\)

\(\overrightarrow{AC}=((t_2-\frac{\alpha}{5}),(\frac{1}{2}+t_2)-2,(-\frac{4}{3}+t_2)-\alpha)\)

Given the area is 21, use the cross product:

\Area = \(\frac{1}{2}\left|(\overrightarrow{AB} \times \overrightarrow{AC})\right| = 21\

This results in:

\[\left|(t_1-t_2)(t_1-\alpha+\frac{4}{3})+(\alpha-(\frac{4}{3}+t_2))((\frac{1}{2}+t_1)-2)\right|=42\]

\Introducing \(u=t_1-v,\ v=2,\) gives: 
\(u(\alpha-\frac{4}{3}+v) - 2(\alpha-\frac{4}{3}+t_2) - (v+\frac{1}{2}-2)u=42\)

Simplify it to: \[42=\left|-\alpha+\frac{8}{3}\right| \] Adjusted: \(\alpha=\pm7\).

4. **Verification**: \(\alpha^2=49\) within [9,9].