Let the position vectors of point A- B C are a - b c respectively- Point D divides line segment BC internally in the ratio 2-1 - Find vector AD-

Positions vectors of A, B amp; C are a , b amp; c .BD : DC = 2:1Position vector of D = 2(c) + 1(b)2+1 = 2c + b3 By triangular ...

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Byju's Answer

Standard XII

Physics

Displacement in 2D Motion

Let the posit...

Question

Let the position vectors of point A, B & C are $\to a$ , $\to b$ & $\to c$ respectively. Point D divides line segment BC internally in the ratio 2:1 . Find vector $- - \to A D$ .

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Solution

Positions vectors of A, B & C are $\to a$ , $\to b$ & $\to c$ .
BD : DC = 2:1
Position vector of D = $\frac{2 (\to c) + 1 (\to b)}{2 + 1}$ = $\frac{2 \to c + \to b}{3}$
By triangular law of vector addition,
$- - \to B A$ + $- - \to A D$ = $- - \to B D$
$- - \to A D$ = $- - \to B D$ - $- - \to B A$
$- - \to A D$ = $\frac{2}{3} (\to c - \to b)$ - $(\to a - \to b)$
$- - \to A D$ = $\frac{2}{3} \to c$ - $\to a$ + $\frac{1}{3} \to b$

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Let the position vectors of point A, B & C are →a , →b & →c respectively. Point D divides line segment BC internally in the ratio 2:1 . Find vector −−→AD.

Let the position vectors of point A, B & C are $\to a$ , $\to b$ & $\to c$ respectively. Point D divides line segment BC internally in the ratio 2:1 . Find vector $- - \to A D$ .