If $A = 2^i + 3^j + 6^k$ and $B = 3^i - 6^j + 2^k$ , then vector perpencicular to both $A$ and $B$ has magnitude $m$ times that of $6^i - 2^j + 3^k$ .Find $m$

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Solution

Let $\to C$ be the vector perpendicular to vector $\to A$ and $\to B$ then $\to C$ can be taken as the cross procuts of $\to A$ and $\to B$
$∴ \to C = \to A \times \to B = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & 3 & 6 3 & - 6 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (3 \times 2 - 6 \times - 6) +^j (6 \times 3 - 2 \times 2) +^k (2 \times - 6 - 3 \times 3)$
$= 42^i + 14^j - 21^k$
From the question,
$\sqrt{{(42)}^{2} + {(14)}^{2} + {(- 21)}^{2}} = m \sqrt{{(6)}^{2} + {(- 2)}^{2} + {(3)}^{2}}$
$\Rightarrow 7 \sqrt{{(6)}^{2} + {(2)}^{2} + {(3)}^{2}} = m \sqrt{{(6)}^{2} + {(2)}^{2} + {(3)}^{2}}$
$∴ m = 7$

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