Question

If $\vec{a} = \vec{i} + \vec{j} + \vec{k}$, $\vec{b} = 4\vec{i} + 3\vec{j} + 4\vec{k}$ and $\vec{c} = \vec{i} + \alpha\vec{j} + \beta\vec{k}$ are linearly dependent vectors and $|\vec{c}| = \sqrt{3}$, then:

• A. $\alpha = 1, \beta = -1$
• B. $\alpha = 1, \beta = 1$
• C. $\alpha = -1, \beta = -1$
• D. $\alpha = 2, \beta = 1$

Hint:

Linear dependency of three vectors in $\mathbb{R}^3$ is equivalent to coplanarity. Always set the determinant to $0$.

Answer 1

The Correct Option is B

Step 1: Concept
Three vectors in three-dimensional space are linearly dependent if and only if they are coplanar. This implies their scalar triple product (or the determinant of their coefficients) is equal to zero.

Step 2: Meaning
We can set up the determinant of the coefficients of $\vec{a}$, $\vec{b}$, and $\vec{c}$ to be zero, and use the magnitude condition $|\vec{c}| = \sqrt{3}$ to solve for $\alpha$ and $\beta$.

Step 3: Analysis
Since the vectors are linearly dependent, the determinant of their coefficients is zero: \[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0 \] Expanding along the first row: \[ 1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0 \] \[ 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0 \] \[ 1 - \beta = 0 \implies \beta = 1 \] Now, using the magnitude of $\vec{c}$: \[ |\vec{c}| = \sqrt{1^2 + \alpha^2 + \beta^2} = \sqrt{3} \] \[ 1 + \alpha^2 + 1 = 3 \implies \alpha^2 = 1 \implies \alpha = \pm 1 \] This gives possible solutions $(\alpha, \beta)$ as $(1, 1)$ or $(-1, 1)$.

Step 4: Conclusion
Comparing with the options, the values $\alpha = 1, \beta = 1$ are correct.

Final Answer: (B)