Consider three vectors A = hat{i} + hat{j} - | vedclass

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(A) Given vectors are $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$.Vector $X = \alpha

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(A) Given vectors are $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$.Vector $X = \alpha A + \beta B = \alpha(\hat{i} + \hat{j} - 2\hat{k}) + \beta(\hat{i} - \hat{j} + \hat{k})$.Simplifying $X$,we get $X = (\alpha + \beta)\hat{i} + (\alpha - \beta)\hat{j} + (-2\alpha + \beta)\hat{k}$.Since $X$ is perpendicular to $C$,their dot product must be zero: $X \cdot C = 0$.$(\alpha + \beta)(2) + (\alpha - \beta)(-3) + (-2\alpha + \beta)(4) = 0$.Expanding the terms: $2\alpha + 2\beta - 3\alpha + 3\beta - 8\alpha + 4\beta = 0$.Combining like terms: $(2 - 3 - 8)\alpha + (2 + 3 + 4)\beta = 0$.$-9\alpha + 9\beta = 0$.$-9\alpha = -9\beta$,which implies $\alpha = \beta$.Therefore,the ratio $\alpha : \beta = 1 : 1$.

Consider three vectors $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$. $A$ vector $X$ of the form $X = \alpha A + \beta B$ (where $\alpha$ and $\beta$ are scalars) is perpendicular to $C$. The ratio of $\alpha$ to $\beta$ is: (in $: 1$)

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