If bar{a}=hat{i}+hat{j}+hat{k},bar{b}=4hat{i} | vedclass

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(A) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are linearly dependent,their scalar triple product must be zero: $\bar{a} \cdot (\bar{b} 	ime

Vedclass · Accepted Answer

$1, 1$

Vedclass · Answer

(A) Since the vectors $\bar{a}$,$\bar{b}$,and $\bar{c}$ are linearly dependent,their scalar triple product must be zero: $\bar{a} \cdot (\bar{b} 	imes \bar{c}) = 0$. This is equivalent to the determinant being zero: $\begin{vmatrix} 1 & 1 & 1 \ 4 & 3 & 4 \ 1 & \alpha & \beta \end{vmatrix} = 0$. Expanding the determinant: $1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0$. $3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0$. $-\beta + 1 = 0 \implies \beta = 1$. Given $|\bar{c}| = \sqrt{3}$,we have $\sqrt{1^2 + \alpha^2 + \beta^2} = \sqrt{3}$. $1 + \alpha^2 + \beta^2 = 3$. Substituting $\beta = 1$: $1 + \alpha^2 + 1 = 3 \implies \alpha^2 = 1 \implies \alpha = \pm 1$. Checking the options,for $\alpha = 1$ and $\beta = 1$,the condition is satisfied. Thus,the correct option is $A$.

If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$,$\bar{b}=4\hat{i}+3\hat{j}+4\hat{k}$,and $\bar{c}=\hat{i}+\alpha\hat{j}+\beta\hat{k}$ are linearly dependent vectors and $|\bar{c}|=\sqrt{3}$,then the values of $\alpha$ and $\beta$ are respectively.

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