If alpha=hat{i}-3 hat{j} and beta=hat{i}+2 ha | vedclass

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(D) Given $\alpha=\hat{i}-3 \hat{j}$ and $\beta=\hat{i}+2 \hat{j}-\hat{k}$.Since $\beta_1$ is parallel to $\alpha$,we can write $\beta_1 = \lambda \al

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None of these

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(D) Given $\alpha=\hat{i}-3 \hat{j}$ and $\beta=\hat{i}+2 \hat{j}-\hat{k}$.Since $\beta_1$ is parallel to $\alpha$,we can write $\beta_1 = \lambda \alpha = \lambda(\hat{i}-3 \hat{j}) = \lambda \hat{i} - 3\lambda \hat{j}$.We have $\beta = \beta_1 + \beta_2$,so $\beta_2 = \beta - \beta_1 = (\hat{i} + 2 \hat{j} - \hat{k}) - (\lambda \hat{i} - 3\lambda \hat{j}) = (1-\lambda)\hat{i} + (2+3\lambda)\hat{j} - \hat{k}$.Since $\beta_2$ is perpendicular to $\alpha$,their dot product is zero: $\beta_2 \cdot \alpha = 0$.$((1-\lambda)\hat{i} + (2+3\lambda)\hat{j} - \hat{k}) \cdot (\hat{i} - 3\hat{j}) = 0$.$(1-\lambda)(1) + (2+3\lambda)(-3) + (-1)(0) = 0$.$1 - \lambda - 6 - 9\lambda = 0$.$-5 - 10\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.Thus,$\beta_1 = -\frac{1}{2}(\hat{i} - 3\hat{j}) = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j}$.Comparing this with the given options,none of the options match the calculated result.

If $\alpha=\hat{i}-3 \hat{j}$ and $\beta=\hat{i}+2 \hat{j}-\hat{k}$,express $\beta$ in the form $\beta=\beta_1+\beta_2$,where $\beta_1$ is parallel to $\alpha$ and $\beta_2$ is perpendicular to $\alpha$. Then $\beta_1$ is given by:

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