Question

If the vector $a =3 \hat{ j }+4 \hat{ k }$ is the sum of two vectors $a _{1}$ and $a _{2}$, vector $a _{1}$ is parallel to $b =\hat{ i }+\hat{ j }$ and vector $a _{2}$ is perpendicular to $b$, then $a _{1}=$

• A. $\frac{1}{2}(\hat{ i }+\hat{ j })$
• B. $\frac{1}{3}(\hat{ i }+\hat{ j })$
• C. $\frac{2}{3}(\hat{ i }+\hat{ j })$
• D. $\frac{3}{2}(\hat{ i }+\hat{ j })$

Answer 1

Answer: (D)

Given,
$a =3 \hat{ j }+4 \hat{ k } \text { and } b =\hat{ i }+\hat{ j }$
$a = a _{1}+ a _{2}$
$a _{1}$ is parallel to $b$.
$\therefore a _{1}=\lambda b =\lambda(\hat{ i }+\hat{ j })$
$a _{2}$ is perpendicular to $b$.
$\therefore a _{2} \cdot b =0$
$\left( a - a _{1}\right) \cdot b =0$
$(3 \hat{ j }+4 \hat{ k }-\lambda(\hat{ i }+\hat{ j })) \cdot(\hat{ i }+\hat{ j })=0$
$(-\lambda \hat{ i }+(3-\lambda) \hat{ j }+4 \hat{ k }) \cdot(\hat{ i }+\hat{ j })=0$
$\Rightarrow -\lambda+3-\lambda=0 \Rightarrow \lambda=\frac{3}{2}$
$\therefore a _{1}=\frac{3}{2}(\hat{ i }+\hat{ j })$