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Equation of a Plane : Vector Form

In a Δ A B C,...

Question

In a $Δ A B C$ , $∠ A = 30^{\circ}, H$ is the orthocentre and $D$ is the mid point of $- - \to B C \cdot$ Segment $H D$ is produced to $T$ such that $H D = D T$ . Then the ratio $\frac{A T}{B C}$ is equal to

1 : 2

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2 : 1

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3 : 2

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2 : 3

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Solution

The correct option is B $2 : 1$
Let $O$ be the circumcentre being selected as origin of reference.
D is the mid-point of BC
$\Rightarrow - - \to O A = \to a, - - \to O B = \to b, - - \to O C = \to c, - - \to O T = \to t \Rightarrow | \to a | = | \to b | = | \to c | = R$ (circumradius)
$\Rightarrow - - \to O A + - - \to O B + - - \to O C = - - \to O A + 2 - - \to O D = - - \to O A + - - \to A H = - - \to O H$
Position vector of $H = \frac{\to a + \to b + \to c + \to t}{2} = \frac{\to b + c}{2}$
$\Rightarrow \frac{\to a + \to t}{2} = \to 0$
$\Rightarrow \to a + \to t = \to 0$
$\Rightarrow \to t = - \to a$
Now, $- - \to A T = - - \to O T - - - \to O A = \to t - \to a = - \to a - \to a = 2 \to a$
since $\to t = - \to a$
$\Rightarrow | A T | = 2 | a | = 2 R$
where $R$ is the circum-radius.
In triangleABC using sine rule $B C = 2 R \cdot sin A$
where $A = 30^{\circ}$ $B C = 2 R sin 30^{\circ} = 2 R \times \frac{1}{2} = R \Rightarrow A T = 2 - - \to B C \Rightarrow \frac{A T}{B C} = \frac{2}{1} = 2 : 1$

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