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

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Aicharya · Accepted Answer

Since $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent. $\begin{aligned} & \left|\begin{array}{lll} \alpha & \beta & 3 \ 0 & 1 & 2 \ 3 & 2 & 1 \end{array}ight|=0 \Rightarrow-3 \alpha+6 \beta-9=0 \ & \Rightarrow \alpha=2 \beta-3 \ & 	ext { Since }|\vec{a}|=\sqrt{14} \Rightarrow \alpha^2+\beta^2+9=14 \ & \Rightarrow(2 \beta-3)^2+\beta^2=5 \ & \Rightarrow(5 \beta-2)(\beta-2)=0 \Rightarrow \beta=2 \quad\left(\beta 
eq \frac{2}{5}ight) \ & \Rightarrow \alpha=1 \Rightarrow \alpha+\beta=3 \end{aligned}$

$\vec{a}=\alpha \hat{i}+\beta \hat{j}+3 \hat{k}, \vec{b}=\hat{j}+2 \hat{k}, \vec{c}=3 \hat{i}+2…

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