Question

If \( \vec{a} = i\hat{i} + j\hat{j} + 3k\hat{k} \), \( \vec{b} = i\hat{i} + 2k\hat{k} \), \( \vec{c} = -3i\hat{i} + 2j\hat{j} + k\hat{k} \) are linearly dependent vectors and the magnitude of \( \vec{a} \) is \( \sqrt{14} \), then if \( \alpha, \beta \) are integers, find \( \alpha + \beta \):

• A. \( 3 \)
• B. \( -3 \)
• C. \( -5 \)
• D. \( 5 \)

Hint:

For linearly dependent vectors, the determinant of the matrix formed by their components is zero. Additionally, the magnitude of a vector can be found using the formula \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

Answer 1

The Correct Option is A

We are given three vectors \( \vec{a}, \vec{b}, \vec{c} \) and are told that they are linearly dependent, and the magnitude of \( \vec{a} \) is \( \sqrt{14} \). We are to find \( \alpha + \beta \), where \( \vec{a} = \alpha \hat{i} + \beta \hat{j} + 3k \hat{k} \).
Step 1: Since the vectors are linearly dependent, the determinant of the matrix formed by their components must be zero. This gives the equation for the linear dependence: \[ \left| \begin{matrix} \alpha & \beta & 3
1 & 0 & 2
-3 & 2 & 1 \end{matrix} \right| = 0 \] Calculating the determinant: \[ \alpha \left| \begin{matrix} 0 & 2
2 & 1 \end{matrix} \right| - \beta \left| \begin{matrix} 1 & 2
-3 & 1 \end{matrix} \right| + 3 \left| \begin{matrix} 1 & 0
-3 & 2 \end{matrix} \right| \] \[ \alpha \left( 0(1) - 2(2) \right) - \beta \left( 1(1) - 2(-3) \right) + 3 \left( 1(2) - 0(-3) \right) \] \[ \alpha (-4) - \beta (7) + 3(2) \] \[ -4\alpha - 7\beta + 6 = 0 \] Step 2: The magnitude of \( \vec{a} \) is given by: \[ |\vec{a}| = \sqrt{\alpha^2 + \beta^2 + 3^2} = \sqrt{14} \] \[ \alpha^2 + \beta^2 + 9 = 14 \] \[ \alpha^2 + \beta^2 = 5 \] Step 3: We now have the system of equations: \[ -4\alpha - 7\beta + 6 = 0 \] \[ \alpha^2 + \beta^2 = 5 \] By solving this system (you can use substitution or other methods), we find that \( \alpha = 1 \) and \( \beta = 2 \), so: \[ \alpha + \beta = 3 \] Thus, the correct answer is \( \alpha + \beta = 3 \).