Prove some facts about vectors in R³

by RoRi

May 3, 2016

Prove that $A,B,C \in \mathbb{R}^3$ are all on the same line if and only if
$(B -A) \times (C -A) = O.$
For two vectors $A,B$ with $A \neq B$ , prove that the line through $A$ and $B$ is the set of all vectors $P$ such that $(P-A) \times (P-B) = O$ .

Proof. Assume $A,B,C \in \mathbb{R}^3$ all lie on a line $L$ , and let $A = (a_1, a_2, a_3), \ B = (b_1, b_2, b_3), \ C = (c_1, c_2, c_3)$ . Then
$L = \{ A + tB \} \quad \implies \quad C = (a_1 + tb_1, a_2 + tb_2, a_3 + tb_3).$

So, we have

$B - A = (b_1 - a_1, b_2 - a_2, b_3 - a_3), \qquad C - A = (tb_1, tb_2, tb_3).$

Thus, $(B-A) \times (C-A)$ is the vector

$\begin{align*} &((b_2 - a_2)tb_3 - (a_3-b_3)tb_2, (a_3 - b_3)tb_1 - (a_1-b_1)tb_3, (a_1 - b_1)tb_2 - (a_2 - b_2)tb_1) \\ &= (0,0,0). \end{align*}$

Conversely, assume $(B-A) \times (C-A) = O$ . From Theorem 13.12(g) we know this is the case if and only if the vectors $B-A$ and $C-A$ are linearly dependent. Hence, there are nonzero $s,t$ such that

$s(B-A) + t(C-A) = O$

Since $t$ is nonzero we may divide by $t$ , and we obtain

$sB - sA + tC - tA = O \quad \implies \quad C = A + \frac{s}{t}(A-B).$

But this means $C$ is on the line passing through $A$ and $B. \qquad \blacksquare$
Proof. If $A \neq B$ , then we know there is a unique line passing through $A$ and $B$ , say
$L = \{ A + t(B-A) \}.$

Thus, any point $P \in L$ is given by

$P = (a_1 + t(b_1 - a_1), a_2 + t(b_2 - a_2), a_3 + t(b_3 - a_3) \}.$

Therefore, we have

$\begin{align*} (P-A) \times (P-B) &= t(b_1 - a_1, b_2 - a_2, b_3 - a_3) \times (1-t)(a_1 - b_1, a_2 - b_2, a_3 - b_3) \\ &= t(B-A) \times (B-A) - t(B-A) \times (B-A) \\ &= O. \qquad \blacksquare \end{align*}$

Related

One comment

April 1, 2023 at 3:48 am

S says:

Line is L=A+t(B-A), not A+tB.

Reply

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply