In the isosceles triangle $A B C, ∣ ∣ \to A B ∣ ∣ = ∣ ∣ \to B C ∣ ∣ = 8,$ a point $E$ divides $A B$ internally in the ratio $1 : 3$ , then the cosine of the angle between $\to C E$ & $\to C A$ is ( where $| \to C A | = 12)$

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

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

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

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

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Solution

The correct option is D $\frac{3 \sqrt{7}}{8}$
Let $B$ be the origin and the position vector of points $A, C$ are $\to a, \to c$ respectively.
$a = c = 8$
$| \to c - \to a | = 12$ ,
$\Rightarrow a^{2} + c^{2} - 2 (\to a . \to c) = 144$ ,
$\Rightarrow \to a . \to c = - 8$
$\to e = \frac{3 \to a + \to b}{4}$

Now, $cos θ = \frac{(\to e - \to c) . (\to a - \to c)}{| \to e - \to c | \times | \to a - \to c |}$
$∴ cos θ = \frac{0.75 \times 64 + 6 + 8 + 64}{\sqrt{(36 + 64 + 12) \times (64 + 64 + 16)}}$

$= \frac{126}{12 \times \sqrt{112}}$

$= \frac{21}{8 \times \sqrt{7}}$

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Similar questions

Q. In the isosceles triangle ABC

∣ ∣ \to A B ∣ ∣ = ∣ ∣ \to B C ∣ ∣ = 8

, a point E divides AB internally in the ratio

1 : 3

, then the cosine of the angle between

\to C E

\to C A

is? (where

∣ ∣ \to C A ∣ ∣ = 12

)

Q. Find the mode of the following data.

\begin{matrix} 2 & 14 & 17 & 11 & 15 7 & 20 & 8 & 17 & 10 8 & 20 & 1 & 8 & 3 \end{matrix}

Q. The point which divides internally , the line joining

(- 3, - 4)

(- 8, 7)

in the ratio

7 : 5

Evaluate :

$(i) \frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9} (i i) \frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5} (i i i) \frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{9} (i v) \frac{3}{8} + \frac{- 5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{- 5}{8} + \frac{- 2}{7}$

Fill in the blanks.

(i) $(\frac{- 3}{17}) + (\frac{- 12}{5}) = (\frac{- 12}{5}) + (. . . . .)$

(ii) $- 9 + \frac{- 21}{8} = (. . . . . .) + (- 9)$

(iii) $(\frac{- 8}{13} + \frac{3}{7}) + (\frac{- 13}{4}) = (. . . . . .) + [\frac{3}{7} + (\frac{- 13}{4})]$

(iv) $- 12 + (\frac{7}{12} + \frac{- 9}{11}) = (- 12 + \frac{7}{12}) + (. . . . .)$

(v) $\frac{19}{- 5} + (\frac{- 3}{11} + \frac{- 7}{8}) = {\frac{19}{- 5} + (. . . . .)} + \frac{- 7}{8}$

(vi) $\frac{- 16}{7} + . . . . . = . . . . . . + \frac{- 16}{7} = \frac{- 16}{7}$

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In the isosceles triangle ABC,∣∣→AB∣∣=∣∣→BC∣∣=8, a point E divides AB internally in the ratio 1:3, then the cosine of the angle between →CE & →CA is ( where |→CA|=12)

In the isosceles triangle $A B C, ∣ ∣ \to A B ∣ ∣ = ∣ ∣ \to B C ∣ ∣ = 8,$ a point $E$ divides $A B$ internally in the ratio $1 : 3$ , then the cosine of the angle between $\to C E$ & $\to C A$ is ( where $| \to C A | = 12)$