If the vector a = 3hat{j} + 4hat{k} is the su | vedclass

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(D) Given, $a = 3\hat{j} + 4\hat{k}$ and $b = \hat{i} + \hat{j}$.Since $a = a_1 + a_2$ and $a_1$ is parallel to $b$, we can write $a_1 = \lambda b = \

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$\frac{3}{2}(\hat{i} + \hat{j})$

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(D) Given, $a = 3\hat{j} + 4\hat{k}$ and $b = \hat{i} + \hat{j}$.Since $a = a_1 + a_2$ and $a_1$ is parallel to $b$, we can write $a_1 = \lambda b = \lambda(\hat{i} + \hat{j})$ for some scalar $\lambda$.Since $a_2$ is perpendicular to $b$, we have $a_2 \cdot b = 0$.Substituting $a_2 = a - a_1$, we get $(a - a_1) \cdot b = 0$, which implies $a \cdot b = a_1 \cdot b$.Calculating $a \cdot b = (3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j}) = 3$.Calculating $a_1 \cdot b = \lambda(\hat{i} + \hat{j}) \cdot (\hat{i} + \hat{j}) = \lambda(1^2 + 1^2) = 2\lambda$.Equating the two, $2\lambda = 3$, so $\lambda = \frac{3}{2}$.Therefore, $a_1 = \frac{3}{2}(\hat{i} + \hat{j})$.

If the vector $a = 3\hat{j} + 4\hat{k}$ is the sum of two vectors $a_1$ and $a_2$, where vector $a_1$ is parallel to $b = \hat{i} + \hat{j}$ and vector $a_2$ is perpendicular to $b$, then $a_1 =$

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