Prove some properties of linearly independent vectors

Let $A,B$ be two linearly independent vectors in $\mathbb{R}^3$ and let

$C = (B \times A) - B.$

Prove that $A$ and $B+C$ are orthogonal.
Prove that $\frac{\pi}{2} < \theta < \pi$ where $\theta$ is defined to be the angle between $B$ and $C$ .
Compute the length of $C$ if $\lVert B \rVert = 1$ and $\lVert B \times A \rVert = 2$ .

Proof. We have
$\begin{align*} C = (B \times A) - B && \implies && B + C &= (B \times A) \\ && \implies && A \cdot (B+C) &= A \cdot (B \times A) \\ && \implies && A \cdot (B+C) &= 0 \end{align*}$

from Theorem 13.12(e) (page 483 of Apostol). Hence, $A$ and $B+C$ are orthogonal $. \qquad \blacksquare$
Proof. Starting with the definition of $C$ we have
$\begin{align*} C = (B \times A) - B && \implies && B + C &= B \times A \\ && \implies && B \cdot (B+C) &= B \cdot (B \times A) \\ && \implies && B^2 + B \cdot C &= 0. \end{align*}$

But, $B \neq 0$ , so $B^2 > 0$ . Therefore, $B \cdot C < 0$ . Hence, we have

$\lVert B \rVert \lVert C \rVert \cos \theta < 0 \quad \implies \quad \cos \theta < 0.$

Therefore,

$\frac{\pi}{2} < \theta < \pi. \qquad \blacksquare$
Again, we start with definition of $C$ ,
$\begin{align*} C = B \times A - B && \implies && B + C &= B \times A \\ && \implies && \lVert B + C \rVert^2 &= \lVert B \times A \rVert^2 \\ && \implies && \lVert B \rVert^2 + \lVert C \rVert^2 + 2B \cdot C &= 4 \\ && \implies && \lVert C \rVert^2 &= 3 - 2 B \cdot C. \end{align*}$

But, from part (b) we know $B \cdot C = -1$ . Hence, $\lVert C \rVert = \sqrt{5}$ .