The vector $\vec a = 3\hat j+4\hat k$ is to be written as the sum of a vector $\vec α$ parallel to vector $\vec b =\hat i +\hat j$ and a vector $\vec β$ perpendicular to b. Then, $\vec α=$

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

Let $\vec α =λ\vec b=λ(\hat i +\hat j)$. Then,

$\vec α +\vec β=\vec a⇒\vec β=\vec a-\vec α =-λ\hat i+(3-λ)\hat j+4\hat k$

Now,

$\vec β⊥\vec b$

$⇒\vec β.\vec b=0$

$⇒\{-λ\hat i+(3-λ)\hat j+4\hat k\}.(\hat i +\hat j)=0$

$⇒-λ+3-λ=0⇒λ=\frac{3}{2}$

$∴\vec α =\frac{3}{2}(\hat i+\hat j)$