Question

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:

• A. 16
• B. 14
• C. 12
• D. 10

Answer 1

The Correct Option is B

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:

\( |\vec{c}| = \sqrt{14} \)

Answer 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).