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: Meaning of linearly dependent.
Three vectors are dependent when they lie in one plane, which means their determinant (scalar triple product) is zero.

Step 2: Set the determinant to zero.
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0 \]

Step 3: Expand it.
Expanding gives \[ (3\beta - 4\alpha) - (4\beta - 4) + (4\alpha - 3) = 0 \]

Step 4: Simplify.
Most terms cancel and we get $1 - \beta = 0$, so $\beta = 1$.

Step 5: Use the length condition.
Given $|\vec c| = \sqrt3$: \[ 1 + \alpha^2 + \beta^2 = 3 \implies 1 + \alpha^2 + 1 = 3 \implies \alpha^2 = 1 \]

Step 6: Match the options.
So $\alpha = \pm 1$ with $\beta = 1$. The option present is \[ \boxed{ \alpha = 1,\ \beta = 1 } \]