The vector b = 3j + 4k is to be written as th | vedclass

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(A) Given $b = 3j + 4k$ and $a = i + j$. Since $b_1$ is parallel to $a$,we can write $b_1 = \lambda a = \lambda(i + j) = \lambda i + \lambda j$. We kn

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

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(A) Given $b = 3j + 4k$ and $a = i + j$. Since $b_1$ is parallel to $a$,we can write $b_1 = \lambda a = \lambda(i + j) = \lambda i + \lambda j$. We know that $b = b_1 + b_2$,where $b_2$ is perpendicular to $a$. Thus,$b_2 = b - b_1 = (0 - \lambda)i + (3 - \lambda)j + 4k = -\lambda i + (3 - \lambda)j + 4k$. Since $b_2 \perp a$,their dot product must be zero: $b_2 \cdot a = 0$. $(-\lambda i + (3 - \lambda)j + 4k) \cdot (i + j) = 0$. $-\lambda + (3 - \lambda) = 0$. $3 - 2\lambda = 0 \Rightarrow \lambda = \frac{3}{2}$. Therefore,$b_1 = \frac{3}{2}(i + j)$.

The vector $b = 3j + 4k$ is to be written as the sum of a vector $b_1$ parallel to $a = i + j$ and a vector $b_2$ perpendicular to $a$. Then $b_1 = $

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