Prove statements about a vector satisfying a given vector equation

Consider unit length, orthogonal vectors $A,B \in \mathbb{R}^3$ , and a vector $P$ such that

$P \times B = A-P.$

Prove the following.

$P$ and $B$ are orthogonal and the length of $P$ is $\frac{1}{2} \sqrt{2}$ .
The vectors $P,B,P \times B$ form a basis for $\mathbb{R}^3$ .
$(P \times B) \times B = -P$ .
$P = \frac{1}{2}A - \frac{1}{2} (A \times B)$ .

Proof. We compute,
$\begin{align*} P \times B = A - P && \implies && B \cdot (P \times B) &= B \cdot A - B \cdot P \\ && \implies && 0 = B \cdot P \end{align*}$

since $B \cdot (P \times B) = 0$ and $B \cdot A = 0$ since $A$ and $B$ are orthogonal by assumption. Thus, $B$ and $P$ are orthogonal. Next,

$\begin{align*} \lVert P \times B \rVert^2 &= \lVert P\rVert^2 \lVert B \rVert^2 - (P \cdot B)^2 \\ &= \lVert P \rVert^2 \end{align*}$

since $\lVert B \rVert^2 = 1$ by hypothesis and $P \cdot B = 0$ . Hence, from the vector equation we have

$\begin{align*} P \times B = A - P && \implies && \lVert P \times \rVert^2 &= \lVert A-P \rVert^2 \\ && \implies && \lVert P \rVert^2 &= \lVert A \rVert^2 - \lVert P \rVert^2 \\ && \implies && 2 \lVert P \rVert^2 &= \lVert A \rVert^2 \\ && \implies && 2 \lVert P \rVert^2 &= 1 \\ && \implies && \lVert P \rVert^2 &= \frac{1}{2} \\ && \implies && \lVert P \rVert &= \frac{\sqrt{2}}{2}. \qquad \blacksquare \end{align*}$
Proof. Since $P$ and $B$ are orthogonal (part (a)), we know the vectors $P,B, P \times B$ are independent. Thus, they form a basis for $\mathbb{R}^3$ since any three independent vectors in $\mathbb{R}^3$ are a basis $. \qquad \blacksquare$
Proof. We compute, the vector $(P \times B) \times B$ is given by
$(p_2 b_3 - p_3 b_2, p_3 b_1 - p_1 b_3, p_1 b_2 - p_2 b_1) \times B.$

Then the three coordinates of this cross product are given by

$\begin{align*} &(p_3 b_1 - p_1 b_3)b_3 - (p_1 b_2 - p_2 b_1)b_2 \\ &(p_1 b_2 - p_2 b_1)b_1 - (p_2 b_3 - p_3 b_2)b_3 \\ &(p_2 b_3 - p_3 b_2)b_2 - (p_3 b_1 - p_1 b_3)b_1. \end{align*}$

Expanding these out we obtain the coordinates

$\begin{align*} &p_3 b_1 b_3 - p_1 b_3^2 - p_1 b_2^2 + p_2 b_1 b_2 \\ &p_1 b_1 b_2 - p_2 b_1^2 - p_2 b_3^2 + p_3 b_2 b_3 \\ &p_2 b_2 b_3 - p_3 b_2^2 - p_3 b_1^2 + p_1 b_1 b_3. \end{align*}$

Since $\lVert B \rVert = 1$ we know $b_1^2 + b_2^2 + b_3^2 = 1$ and since $P \cdot B = 0$ we know $p_1 b_1 + p_2 b_2 + p_3 b_3 = 0$ . So, simplifying the expressions above, for each of the coordinates we have

$\begin{align*} p_3 b_1 b_3 - p_1 b_3^2 - p_1 b_2^2 + p_2 b_1 b_2 &= p_3 b_1 b_3 - p_1 (1-b_1^2) + p_2 b_1 b_2 \\ &= p_3 b_1 b_3 - p_1 + p_1 b_1^2 + p_2 b_1 b_2 \\ &= b_1(p_3 b_3 + p_1 b_1 + p_2 b_2) - p_1 \\ &= -p_1 \\[9pt] p_1 b_1 b_2 - p_2 b_1^2 - p_2 b_3^2 + p_3 b_2 b_3 &= p_1 b_1 b_2 - p_2 (1-b_2^2) + p_3 b_2 b_3 \\ &= p_1 b_1 b_2 - p_2 + p_2 b_2^2 + p_3 b_2 b_3 \\ &= b_2(p_1 b_1 + p_2 b_2 + p_3 b_3) - p_2 \\ &= -p_2 \\[9pt] p_2 b_2 b_3 - p_3 b_2^2 - p_3 b_1^2 + p_1 b_1 b_3 &= p_2 b_2 b_3 - p_3 (1-b_3^2) + p_1 b_1 b_3 \\ &= p_2 b_2 b_3 - p_3 + p_3 b_3^2 + p_1 b_1 b_3 \\ &= b_3 (p_2 b_2 + p_3 b_3 + p_1 b_3) - p_3 \\ &= -p_3. \end{align*}$

Hence, we indeed have

$(P \times B) \times B = (-p_1, -p_2, -p_3) = -P. \qquad \blacksquare$
Proof. We compute
$\begin{align*} (P \times B) \times B = -P && \implies && (A-P) \times B &= -P \\ && \implies && (A \times B) - (P \times B) &= -P \\ && \implies && (A \times B) - A + P &= -P \\ && \implies && -2P &= (A \times B) - A \\ && \implies && P &= \frac{1}{2}A - \frac{1}{2}(A \times B). \qquad \blacksquare \end{align*}$

One comment

February 3, 2020 at 6:16 pm

Eiji says:

For 15(a),you calculated ∣∣A-P∣∣² wrong. ∣∣A-P∣∣² should be as follows:
∣∣A-P∣∣² = ∣∣A∣∣² – 2A⋅P + ∣∣P∣∣².

Then you should get ∣∣A∣∣² – 2A⋅P + ∣∣P∣∣² = ∣∣P∣∣². So, A⋅P = ½.

Now you can go to the original equation PXB = A-P, dot product both sides by P and get P⋅(PXB) = P⋅(A-P). (=>) 0 = P.A – P⋅P. (=>) ½ = P⋅P. And you have P = √2/2

Stumbling Robot

Prove statements about a vector satisfying a given vector equation

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