JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Vector Algebra

The projection of $\vec{b}$ on $\vec{a}$ being $\vec{a}$ itself means $\frac{\vec{a}\cdot \vec{b}}{|\vec{a}{|}^2}\vec{a}=\vec{a}$, so $\vec{a}\cdot \vec{b}=|\vec{a}{|}^2=1+1+2=4$. With $\vec{b}=(b_1,b_2,\sqrt{2})$ this reads $b_1+b_2+2=4$, i.e. $b_1+b_2=2$. Next, $\vec{a}+\vec{b}$ perpendicular to $\vec{c}=(5,1,\sqrt{2})$ gives $(1+b_1)5+(1+b_2)1+(\sqrt{2}+\sqrt{2})\sqrt{2}=0$, which simplifies to $5b_1+b_2=-10$. Subtracting the first relation, $4b_1=-12$, so $b_1=-3$ and $b_2=5$, and $|\vec{b}|=\sqrt{(-3{)}^2+5^2+(\sqrt{2}{)}^2}=\sqrt{9+25+2}=6$.