First slide

Algebra of vectors

Question

If H is the orthocentre and D is the midpoint of BC of the $∆ A B C$ and $∠ A = 30^{°}$ . Segment HD is produced to T such that HD= DT then the length AT is equal to

Difficult

A

2BC
B

3BC
C

$\frac{3}{2} BC$
D

none of these

Solution

Let the origin of reference be the circumcentre of the triangle
$Let \vec{O A} = \vec{a}, \vec{O B} = \vec{b}, \vec{O C} = \vec{c} and \vec{O T} = \vec{t}$
$Then | \vec{a} | = | \vec{b} | = | \vec{c} | = R (circumradius)$
$Again \vec{OA} + \vec{OB} + \vec{OC} = \vec{OA} + 2 \vec{OD}$
$= \vec{OA} + \vec{AH} = \vec{OH}$
Therefore, the P.V. of H is $\vec{a} + \vec{b} + \vec{c}$ Since D is the midpoint of HT, we have
$\begin{array}{l} \frac{\vec{a} + \vec{b} + \vec{c} + \vec{t}}{2} = \frac{\vec{b} + \vec{c}}{2} \Rightarrow \vec{t} = - \vec{a} \\ ∴ \vec{AT} = - 2 \vec{a} or \vec{AT} = | - 2 \vec{a} | = 2 | \vec{a} | = 2 R . \end{array}$
But BC =2R sin A = R;
therefore, AT=2BC

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