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Addition of Vectors

ABC is any tr...

Question

$A B C$ is any triangle and $D, E, F$ are the middle points of its sides $B C, C A, A B$ respectively. Express the vectors $- - \to C F$ and $- - \to B E$ as linear combination of the vectors $- - \to A B$ and $- - \to A C$

A

- - \to B E = \frac{1}{2} - - \to A C - - - \to A B; - - \to C F = \frac{1}{2} - - \to A B - - - \to A C

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B

- - \to B E = \frac{1}{2} - - \to A C + - - \to A B; - - \to C F = \frac{1}{2} - - \to A B + - - \to A C

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C

- - \to C F = \frac{1}{2} - - \to A C - - - \to A B; - - \to B E = \frac{1}{2} - - \to A B - - - \to A C

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D

- - \to C F = \frac{1}{2} - - \to A C + - - \to A B; - - \to B E = \frac{1}{2} - - \to A B + - - \to A C

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Solution

The correct option is A $- - \to B E = \frac{1}{2} - - \to A C - - - \to A B; - - \to C F = \frac{1}{2} - - \to A B - - - \to A C$
$- - \to C A + - - \to A F = - - \to C F$
$\Rightarrow - - - \to A C + \frac{1}{2} - - \to A B = - - \to C F$
Similarly $- - \to A E + - - \to B A = - - \to B E$
$\Rightarrow \frac{1}{2} - - \to A C - - - \to A B = - - \to B E$

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Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

can be written as product of two determinants.

e

e_{1}

are the eccentricities of a hyperbola and its conjugate, then

\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}}

\frac{1}{12}

\frac{1}{15}

\frac{1}{17}

, is the greatest fraction

Q. If P and Q are respected by the complex numbers

z_{1}

z_{2}

| \frac{1}{z_{1}} + \frac{1}{z_{2}} | = | \frac{1}{z_{1}} - \frac{1}{z_{2}} |

then the circumcentre of

Δ O P Q

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α

β

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f (x) = x^{2} - p x + q

, then the polynomial having

\frac{1}{α}

\frac{1}{β}

as its zeroes is __________.

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