Let a= i+5 j+ k, b= i+3 j+ k, c=- i+2 j-3 k, with | b c|=5√3 and a b.

Question

Let $\vec a=\hat i+5\hat j+\alpha\hat k$, $\vec b=\hat i+3\hat j+\beta\hat k$, $\vec c=-\hat i+2\hat j-3\hat k$, with $|\vec b	imes\vec c|=5\sqrt3$ and $\vec a\perp\vec b$. The greatest value of $|\vec a|^2$ is \_\_\_\_\_.

Answer: 90

Accepted Answer

$\vec a\cdot\vec b=0$: $1+15+\alpha\beta=0 \implies \alpha\beta=-16$ \ldots  (i)

$$\vec b\times\vec c = \begin{vmatrix}\hat i&\hat j&\hat k\1&3&\beta\-1&2&-3\end{vmatrix} = (-9-2\beta,\ 3-\beta,\ 5)$$

$|\vec b\times\vec c|^2=75$: $(9+2\beta)^2+(3-\beta)^2+25=75 \implies \beta^2+6\beta+8=0 \implies \beta=-2,-4$

Using (i): $\beta=-2\Rightarrow\alpha=8$; $\beta=-4\Rightarrow\alpha=4$. Maximum $|\vec a|^2=1+25+\alpha^2$ occurs at $\alpha=8$:
$$|\vec a|^2 = 1+25+64=90$$

Solutions · 1