Let a= i+ j+√2 k, b=b_1 i+b_2 j+√2 k, c=5 i+ j+√2 k, with the projection vector of b on a …

Question

Let $\vec a=\hat i+\hat j+\sqrt2\hat k$, $\vec b=b_1\hat i+b_2\hat j+\sqrt2\hat k$, $\vec c=5\hat i+\hat j+\sqrt2\hat k$, with the projection vector of $\vec b$ on $\vec a$ equal to $\vec a$. If $\vec a+\vec b\perp\vec c$, then $|\vec b|$ is:

Accepted Answer

Projection of $\vec b$ on $\vec a$ equals $\vec a$ means $\dfrac{\vec b\cdot\vec a}{|\vec a|^2}\vec a=\vec a \Rightarrow \vec b\cdot\vec a=|\vec a|^2=4$:
$$b_1+b_2+2=4 \implies b_1+b_2=2 \quad (1)$$

$(\vec a+\vec b)\cdot\vec c=0$: $\vec a\cdot\vec c+\vec b\cdot\vec c=0$. $\vec a\cdot\vec c=5+1+2=8$. $\vec b\cdot\vec c=5b_1+b_2+2$:
$$8+5b_1+b_2+2=0 \implies 5b_1+b_2=-10 \quad (2)$$

Solving (1),(2): $b_1=-3,\ b_2=5$. $\vec b=-3\hat i+5\hat j+\sqrt2\hat k$.

$$|\vec b| = \sqrt{9+25+2}=\sqrt{36}=6$$

Solutions · 1