First slide

Cross product of vectors or vector product of vectors

Question

$If \vec{a}, \vec{b}, and \vec{c} are unit vectors such that \vec{a} + 2 \vec{b} + 2 \vec{c} = \vec{0}, then | \vec{a} \times \vec{c} | is equal to$

Moderate

A

$\frac{\sqrt{15}}{4}$
B

$\frac{\sqrt{15}}{16}$
C

$\frac{15}{16}$
D

$\frac{15}{4}$

Solution

$2 \vec{b} = - \vec{a} - 2 \vec{c}$
squaring both sides
$4 {\vec{b}}^{2} = {\vec{a}}^{2} + 4 {\vec{c}}^{2} + 4 \vec{a} . \vec{c}$
$4 = 1 + 4 + 4 \vec{a} . \vec{c}$
$\Rightarrow \vec{a} . \vec{c} = - \frac{1}{4}$
${|\bar{a} \times \bar{c}|}^{2} = {\bar{a}}^{2} {\bar{c}}^{2} - {(\bar{a} . \bar{c})}^{2}$
$= 1 - \frac{1}{16} = \frac{15}{16}$

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