First slide

Scalar triple product of vectors

Question

If vectors $\vec{a}, \vec{b} and \vec{c}$ are non-coplanar and l, m and n are distinct scalars, then $[(l \vec{a} + m \vec{b} + n \vec{c}) (l \vec{b} + m \vec{c} + n \vec{a}) (l \vec{c} + m \vec{a} + n \vec{b})] = 0$ implies

Moderate

A

l+m+n=0
B

roots of the equation lx²+ mx + n=0 are real
C

l²+m²+n²=0
D

l³+m³+n³=3lmn

Solution

$\begin{array}{l} {\vec{V}}_{1} = l \vec{a} + m \vec{b} + n \vec{c} \\ {\vec{V}}_{2} = n \vec{a} + l \vec{b} + m \vec{c} \\ {\vec{V}}_{3} = m \vec{a} + n \vec{b} + l \vec{c} \end{array}\} \begin{matrix} when \vec{a}, \vec{b} and \vec{c} \\ are non-coplanar. \end{matrix}$
Therefore,
$\begin{array}{l} [\begin{array}{lll} {\vec{V}}_{1} {\vec{V}}_{2} {\vec{V}}_{3}] \end{array}] = |\begin{array}{ccc} l & m & n \\ n & l & m \\ m & n & l \end{array}| = 0 \\ or (l + m + n) [(l - m)^{2} + (m - n)^{2} + (n - l)^{2}] = 0 \\ or l + m + n = 0 - - - (i) \end{array}$
Obviously, l²+ mx + n=0 is satisfied by x=1 due to (i).
$\begin{array}{l} l^{3} + m^{3} + n^{3} = 3 lmn \\ (l + m + n) (l^{2} + m^{2} + n^{2} - lm - mn - \ln) = 0 \end{array}$
which is true

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