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Applications of Cross Product

Find unit vec...

Question

Find unit vectors perpendicular to the plane of the vectors,
$\to A =^i - 2^j +^k$ and $\to B = 2^i -^k$ .

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Solution

$\begin{matrix} \to A =^i - 2^j +^k \to B = 2^i -^k \to A \times \to B = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 2 & 1 2 & 0 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i ∣ ∣ ∣ \begin{matrix} - 2 & 1 0 & - 1 \end{matrix} ∣ ∣ ∣ -^j ∣ ∣ ∣ \begin{matrix} 1 & 1 2 & - 1 \end{matrix} ∣ ∣ ∣ +^k ∣ ∣ ∣ \begin{matrix} 1 & - 2 2 & 0 \end{matrix} ∣ ∣ ∣ =^i [2 - 0] -^j [- 1 - 2] +^k [0 + 4] = 2^i + 3^j + 4^k ∣ ∣ \to A \times \to B ∣ ∣ = \pm \sqrt{2^{2} + 3^{2} + 4^{2}} = \pm \sqrt{29} ∴ u n i t v e c t o r = \frac{\to A \times \to B}{∣ ∣ \to A \times \to B ∣ ∣} = \pm \frac{1}{\sqrt{29}} (2^i + 3^j + 4^k) A n s . \frac{2^i + 3^j + 4^k}{\sqrt{29}}, \frac{- (2^i + 3^j + 4^k)}{\sqrt{29}} \end{matrix}$

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\to a =^i +^j -^k, \to b = -^i + 2^j +^k

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