If vec{a}=alpha hat{i}+beta hat{j}+3 hat{k},v | vedclass

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(A) Since the vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent,their scalar triple product must be zero: $\begin{vmatrix} \alpha & \beta & 3

Vedclass · Accepted Answer

$3$

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(A) Since the vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent,their scalar triple product must be zero: $\begin{vmatrix} \alpha & \beta & 3 \ 0 & 1 & 2 \ 3 & 2 & 1 \end{vmatrix} = 0$ Expanding along the first row: $\alpha(1 - 4) - \beta(0 - 6) + 3(0 - 3) = 0$ $-3\alpha + 6\beta - 9 = 0$ Dividing by $-3$: $\alpha - 2\beta + 3 = 0 \Rightarrow \alpha = 2\beta - 3$ Given that the magnitude $|\vec{a}| = \sqrt{14}$,we have: $\alpha^2 + \beta^2 + 3^2 = 14$ $\alpha^2 + \beta^2 = 5$ Substituting $\alpha = 2\beta - 3$: $(2\beta - 3)^2 + \beta^2 = 5$ $4\beta^2 - 12\beta + 9 + \beta^2 = 5$ $5\beta^2 - 12\beta + 4 = 0$ Factoring the quadratic equation: $(5\beta - 2)(\beta - 2) = 0$ Since $\beta$ is an integer,we take $\beta = 2$. Then $\alpha = 2(2) - 3 = 1$. Therefore,$\alpha + \beta = 1 + 2 = 3$.

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

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