Let three vectors \( \vec{a} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k} \),
\( \vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} \),
\( \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \) from a triangle such that \( \vec{c} = \vec{a} - \vec{b} \) and the area of the triangle is \( 5 \sqrt{6} \). If \(\alpha\) is a positive real number, then \( |\vec{c}|^2 \) is:
\( \vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} \),
\( \vec{c} = x \hat{i} + y \hat{j} + z \hat{k} \) from a triangle such that \( \vec{c} = \vec{a} - \vec{b} \) and the area of the triangle is \( 5 \sqrt{6} \). If \(\alpha\) is a positive real number, then \( |\vec{c}|^2 \) is:
- 16
- 14
- 12
- 10
The Correct Option is B
Approach Solution - 1
Given:
\( \vec{c} = \vec{a} - \vec{b} \)
Let \( \vec{a} = (\alpha, 4, 2) \) and \( \vec{b} = (5, 3, 4) \). Then, \( \vec{c} = (x, y, z) = (\alpha - 5, 1, -2) \)
Hence, \( x = \alpha - 5, \, y = 1, \, z = -2 \quad \ldots (1) \)
The area of the triangle is given as \( 5\sqrt{6} \).
Using the formula for the area of a triangle formed by two vectors:
\( \frac{1}{2} |\vec{a} \times \vec{c}| = 5\sqrt{6} \)
Therefore,
\( |\vec{a} \times \vec{c}| = 10\sqrt{6} \)
Now, compute the cross product:
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ x & 1 & -2 \end{vmatrix} \]
\( = \hat{i}(4(-2) - 2(1)) - \hat{j}(\alpha(-2) - 2x) + \hat{k}(\alpha(1) - 4x) \)
\( = \hat{i}(-10) - \hat{j}(-2\alpha - 2x) + \hat{k}(\alpha - 4x) \)
Taking magnitude:
\( |(-10\hat{i} - \hat{j}(-2\alpha - 2x) + \hat{k}(\alpha - 4x))| = 10\sqrt{6} \)
Thus,
\( (2\alpha + 2x - 10)^2 + (\alpha - 4x + 20)^2 = 500 \)
Simplifying:
\( (4\alpha - 10)^2 + (20 - 3\alpha)^2 = 500 \)
\( 25\alpha^2 - 80\alpha - 120\alpha = 0 \)
\( \alpha (25\alpha - 200) = 0 \)
\( \alpha = 8 \) (since \( \alpha \) is positive)
Substituting back into (1):
\( x = \alpha - 5 = 3 \)
Now, the magnitude of \( \vec{c} \) is:
\( |\vec{c}|^2 = x^2 + y^2 + z^2 \)
\( = 9 + 1 + 4 = 14 \)
Hence, the final value is:
\( |\vec{c}| = \sqrt{14} \)
Approach Solution -2
Step 1: Expression for \(\vec{c}\)
The vector \(\vec{c}\) is given as:
\(\vec{c} = \vec{a} - \vec{b}\).
Substitute \(\vec{a} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}\) and \(\vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}\):
\(\vec{c} = (\alpha - 5)\hat{i} + (4 - 3)\hat{j} + (2 - 4)\hat{k}\).
Thus:
\(\vec{c} = (\alpha - 5)\hat{i} + \hat{j} - 2\hat{k}\).
Step 2: Area of the triangle
The area of the triangle is given as:
\(\text{Area} = \frac{1}{2} |\vec{a} \times \vec{c}|.\)
Substitute \(\text{Area} = 5\sqrt{6}\):
\(\frac{1}{2} |\vec{a} \times \vec{c}| = 5\sqrt{6}.\)
\(|\vec{a} \times \vec{c}| = 10\sqrt{6}.\)
Step 3: Cross product \(\vec{a} \times \vec{c}\)
The cross product \(\vec{a} \times \vec{c}\) is given by:
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ \alpha - 5 & 1 & -2 \end{vmatrix}. \]
Expanding the determinant:
\(\vec{a} \times \vec{c} = \hat{i} \begin{vmatrix} 4 & 2 \\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} \alpha & 2 \\ \alpha - 5 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} \alpha & 4 \\ \alpha - 5 & 1 \end{vmatrix}. \)
Calculate each minor:
1. For \(\hat{i}\): \(\begin{vmatrix} 4 & 2 \\ 1 & -2 \end{vmatrix} = (4)(-2) - (2)(1) = -8 - 2 = -10.\)
2. For \(\hat{j}\): \(\begin{vmatrix} \alpha & 2 \\ \alpha - 5 & -2 \end{vmatrix} = (\alpha)(-2) - (\alpha - 5)(2) = -2\alpha - 2\alpha + 10 = -4\alpha + 10.\)
3. For \(\hat{k}\): \(\begin{vmatrix} \alpha & 4 \\ \alpha - 5 & 1 \end{vmatrix} = (\alpha)(1) - (\alpha - 5)(4) = \alpha - 4\alpha + 20 = -3\alpha + 20.\)
Thus:
\(\vec{a} \times \vec{c} = -10\hat{i} - (-4\alpha + 10)\hat{j} + (-3\alpha + 20)\hat{k}.\)
\(\vec{a} \times \vec{c} = -10\hat{i} + (4\alpha - 10)\hat{j} + (-3\alpha + 20)\hat{k}.\)
Step 4: Magnitude of \(\vec{a} \times \vec{c}\)
The magnitude is:
\(|\vec{a} \times \vec{c}| = \sqrt{(-10)^2 + (4\alpha - 10)^2 + (-3\alpha + 20)^2}.\)
\(|\vec{a} \times \vec{c}| = \sqrt{100 + (16\alpha^2 - 80\alpha + 100) + (9\alpha^2 - 120\alpha + 400)}.\)
\(|\vec{a} \times \vec{c}| = \sqrt{25\alpha^2 - 200\alpha + 600}.\)
Set \(|\vec{a} \times \vec{c}| = 10\sqrt{6}\):
\(\sqrt{25\alpha^2 - 200\alpha + 600} = 10\sqrt{6}.\)
Square both sides:
\(25\alpha^2 - 200\alpha + 600 = 600.\)
\(25\alpha(\alpha - 8) = 0.\)
Since \(\alpha > 0\), \(\alpha = 8.\)
Step 5: Calculate \(|\vec{c}|^2\)
Substitute \(\alpha = 8\) into \(\vec{c}\):
\(\vec{c} = 3\hat{i} + \hat{j} - 2\hat{k}.\)
The magnitude squared is:
\(|\vec{c}|^2 = 3^2 + (1)^2 + (-2)^2.\)
\(|\vec{c}|^2 = 9 + 1 + 4 = 14.\)
Final Answer: Option (2).
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