First slide

Dot product or scalar product of vectors

Question

$In the isosceles triangle \underline{ABC,} | \vec{A B} | = | \vec{B C} | = 8, a point E divides AB internally in$ $the ratio 1 : 3, then the cosine of the angle between \vec{C E} and \vec{C A} is : (where | \vec{C A} | = 12)$

Difficult

A

$- \frac{3 \sqrt{7}}{8}$
B

$\frac{3 \sqrt{8}}{17}$
C

$\frac{3 \sqrt{7}}{8}$
D

$- \frac{3 \sqrt{8}}{17}$

Solution

$Given | \vec{b} | = | \vec{b} - \vec{c} | = 8 and | \vec{c} | = 12$
$| \vec{b} |^{2} = | \vec{b} - \vec{c} |^{2} | \vec{b} |^{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2} - 2 |\vec{b} \cdot \vec{c}| \vec{b} \cdot \vec{c} = \frac{144}{2} = 72$
${(\vec{c} - \frac{\vec{b}}{4})}^{2} = {|\vec{c}|}^{2} - 2 \vec{c} \cdot \frac{\vec{b}}{4} + \frac{{\vec{b}}^{2}}{16} = 144 - \frac{72}{4} + \frac{64}{16} = 112 |\vec{c} - \frac{\vec{b}}{4}| = \sqrt{112}$
$\vec{C E} = \vec{O E} - \vec{O C} = \frac{\vec{b}}{4} - \vec{c}$ $\vec{C A} = - \vec{c}$
$\cos θ = \frac{\vec{c} (\vec{c} - \frac{\vec{b}}{4})}{| \vec{c} | |\vec{c} - \frac{\vec{b}}{4}|} = \frac{\vec{c^{2}} - \frac{\vec{c} \cdot \vec{b}}{4}}{12 |\vec{c} - \frac{\vec{b}}{4}|}$
$= \frac{144 - 18}{12 \sqrt{112}} = \frac{126}{12 \times 4 \sqrt{7}}$
$= \frac{3 \sqrt{7}}{8}$

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