JEE Main 2019 (Online) 9th January Evening ShiftQuestion 1
Let $\vec a=\hat i+\hat j+\sqrt2\hat k$, $\vec b=b_1\hat i+b_2\hat j+\sqrt2\hat k$ and $\vec c=5\hat i+\hat j+\sqrt2\hat k$ be three vectors such that the projection vector of $\vec b$ on $\vec a$ is $|\vec a|$. If $\vec a+\vec b$ is perpendicular to $\vec c$, then $|\vec b|$ is equal to :
Vector Algebra
Solution
Projection of $\vec b$ on $\vec a$ equal to $|\vec a|$: $\vec a\cdot\vec b=|\vec a|^2=4$, so $b_1+b_2=2$. $(\vec a+\vec b)\cdot\vec c=0$: $5b_1+b_2=-10$. Then $b_1=-3$, $b_2=5$, $|\vec b|^2=9+25+2=36$.