First slide

Vectors

Question

Unit vector perpendicular to vector $A = - 3 \hat{i} - 2 \hat{j} - 3 \hat{k} and B = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ both is

Moderate

A

$\frac{3 \hat{j} - 2 \hat{k}}{\sqrt{13}}$
B

$\frac{3 \hat{k} - 2 \hat{j}}{\sqrt{13}}$
C

$\frac{- \hat{j} + 2 \hat{k}}{\sqrt{13}}$
D

$\frac{\hat{i} + 3 \hat{j} - \hat{k}}{\sqrt{13}}$

Solution

Unit vector is given by $\hat{n} = \frac{A \times B}{| A \times B |}$
$A \times B = |\begin{array}{r} \hat{i} & \hat{j} & \hat{k} \\ - 3 & - 2 & - 3 \\ 2 & 4 & 6 \end{array}|$
$= \hat{i} (- 12 + 12) - \hat{j} (- 18 + 6) + ∣ \hat{k} (- 12 + 4) = 12 \hat{j} - 8 \hat{k}$
$| A \times B | = \sqrt{(12)^{2} + (- 8)^{2}} = \sqrt{208} = 4 \sqrt{13}$
$∴ \hat{n} = \frac{12 \hat{j} - 8 \hat{k}}{4 \sqrt{13}} = \frac{3 \hat{j} - 2 \hat{k}}{\sqrt{13}}$

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