© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the given data
• One vertex of the triangle is given as A(0, 2, α).
• The other two vertices, B and C, lie on the line
\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3} .
• The length of side BC is 2\sqrt{21} .
• The area of the triangle ABC is 21 square units.
• We need to find \alpha^2 for integer α.
Step 2: Write a point on the line and its direction vector
Any point on the line can be found by setting a parameter t :
From \frac{x+\alpha}{5} = t , we get x = 5t - \alpha.
From \frac{y-1}{2} = t , we get y = 2t + 1.
From \frac{z+4}{3} = t , we get z = 3t - 4.
Thus, a generic point on the line is (\,5t - \alpha,\, 2t + 1,\, 3t - 4 ) .
A convenient point on the line (for instance, at t = 0 ) is
P_0(-\alpha,\, 1,\, -4).
The direction vector of this line is
\vec{d} = (5,\, 2,\, 3).
Step 3: Use the area condition to find the distance of A from the line
The area of triangle ABC can be viewed as
\text{Area} = \tfrac{1}{2} \times (\text{base}) \times (\text{height}).
Here, BC is the base of length 2\sqrt{21} , and the “height” is the perpendicular distance from A to the line containing BC.
Therefore:
21 \;=\; \tfrac{1}{2} \times 2\sqrt{21} \times \bigl(\text{distance from A to line}\bigr).
So, the distance from A to the line is
\text{distance} \;=\; \dfrac{21}{\sqrt{21}} \;=\; \sqrt{21}.
Step 4: Compute the perpendicular distance from A to the line
The distance from a point \bigl( x_1,\, y_1,\, z_1 \bigr) to a line passing through a point \bigl( x_0,\, y_0,\, z_0 \bigr) with direction vector \vec{d} is
\text{Distance} \;=\; \dfrac{\bigl|\bigl(\vec{AP_0}\bigr) \times \vec{d}\bigr|}{\lVert \vec{d}\rVert},
where \vec{AP_0} is the vector from P_0 (a point on the line) to A.
In our problem:
P_0 = (-\alpha,\, 1,\, -4).
A = (0,\, 2,\, \alpha).
\vec{AP_0} = \bigl(0 - (-\alpha),\; 2 - 1,\; \alpha - (-4)\bigr)
= (\alpha,\, 1,\, \alpha + 4).
\vec{d} = (5,\, 2,\, 3).
\lVert \vec{d}\rVert = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{38}.
Thus,
\bigl|\bigl(\vec{AP_0}\bigr) \times \vec{d}\bigr|
\;=\; \text{(distance)} \times \lVert \vec{d}\rVert
\;=\; \sqrt{21} \times \sqrt{38}
\;=\; \sqrt{21 \times 38}
\;=\; \sqrt{798}.
Step 5: Evaluate the cross product and form the equation
Compute
\bigl(\vec{AP_0}\bigr) \times \vec{d}
=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\alpha & 1 & \alpha + 4 \\
5 & 2 & 3
\end{vmatrix}.
Expanding this determinant:
\bigl(\vec{AP_0}\bigr) \times \vec{d}
= \mathbf{i}\bigl(1 \cdot 3 \;-\; (\alpha+4)\cdot 2\bigr)
-\mathbf{j}\bigl(\alpha \cdot 3 \;-\; 5(\alpha+4)\bigr)
+\mathbf{k}\bigl(\alpha \cdot 2 \;-\; 5 \cdot 1\bigr).
Simplify each component:
\mathbf{i} -component = 3 - 2(\alpha + 4) = 3 - 2\alpha - 8 = -2\alpha - 5.
\mathbf{j} -component = -\bigl(3\alpha - 5\alpha - 20\bigr)
= -\bigl(-2\alpha - 20\bigr)
= 2\alpha + 20.
\mathbf{k} -component = 2\alpha - 5.
So,
\bigl(\vec{AP_0}\bigr) \times \vec{d} = \bigl(-2\alpha - 5,\; 2\alpha + 20,\; 2\alpha - 5\bigr).
Next, the magnitude is
\bigl|\bigl(\vec{AP_0}\bigr) \times \vec{d}\bigr|
= \sqrt{\bigl(-2\alpha - 5\bigr)^2 + \bigl(2\alpha + 20\bigr)^2 + \bigl(2\alpha - 5\bigr)^2}.
Step 6: Impose the condition for the required distance
We have
\bigl|\bigl(\vec{AP_0}\bigr) \times \vec{d}\bigr| = \sqrt{798}.
Thus,
(-2\alpha - 5)^2 + (2\alpha + 20)^2 + (2\alpha - 5)^2 = 798.
Expand and combine like terms:
(-2\alpha - 5)^2 = 4\alpha^2 + 20\alpha + 25,
(2\alpha + 20)^2 = 4\alpha^2 + 80\alpha + 400,
(2\alpha - 5)^2 = 4\alpha^2 - 20\alpha + 25.
Summing up gives
12\alpha^2 + 80\alpha + (25 + 400 + 25) = 12\alpha^2 + 80\alpha + 450.
So the equation is
12\alpha^2 + 80\alpha + 450 = 798 \quad\Longrightarrow\quad 12\alpha^2 + 80\alpha - 348 = 0.
Step 7: Solve the quadratic equation for α
12\alpha^2 + 80\alpha - 348 = 0.
Divide throughout by 4 to make it simpler (or use the quadratic formula directly).
Either approach will show that the integer root is
\alpha = 3.
(Any other solution is non-integer and hence not valid under the given conditions.)
Step 8: Conclude the value of α²
Since \alpha = 3 ,
\alpha^2 = 9.
Thus, the required value of \alpha^2 is
9.
