Home » Blog » Prove some facts about vectors in R3

Prove some facts about vectors in R3

  1. Prove that A,B,C \in \mathbb{R}^3 are all on the same line if and only if

        \[ (B -A) \times (C -A) = O. \]

  2. 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.

  1. 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

  2. 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*}

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