Let ABCD is a parallelogram- position vector of points A-C- and D are a - c and d respectively-if E divides line segment AB internally in the ratio 3-2 then find the vector DE-

Let B be origin vector ∴ #160; E = 3(0̅)+2(a⃗)3+2=2a⃗5 D⃗E⃗=E⃗ D⃗ D⃗E⃗=2a⃗5 d⃗ #160; #160; #160; #160; =15(2a⃗ 5d⃗) ...

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Standard XII

Mathematics

Distance Formula

let ABCD is a...

Question

let ABCD is a parallelogram. position vector of points A,C, and D are $\to a, \to c$ and $\to d$ respectively.
if E divides line segment AB internally in the ratio 3:2 then find the vector $- - \to D E$ .

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Solution

Let $B$ be origin vector
$∴ E = \frac{3 (¯ 0) + 2 (\to a)}{3 + 2} = \frac{2 \to a}{5}$
$\to D E = \to E - \to D$
$\to D E = \frac{2 \to a}{5} - \to d$
$= \frac{1}{5} (2 \to a - 5 \to d)$

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let ABCD is a parallelogram. position vector of points A,C, and D are →a,→c and →d respectively.if E divides line segment AB internally in the ratio 3:2 then find the vector −−→DE.

Let B be origin vector∴E=3(¯0)+2(→a)3+2=2→a5→DE=→E−→D→DE=2→a5−→d =15(2→a−5→d)

let ABCD is a parallelogram. position vector of points A,C, and D are $\to a, \to c$ and $\to d$ respectively.
if E divides line segment AB internally in the ratio 3:2 then find the vector $- - \to D E$ .

Let $B$ be origin vector
$∴ E = \frac{3 (¯ 0) + 2 (\to a)}{3 + 2} = \frac{2 \to a}{5}$
$\to D E = \to E - \to D$
$\to D E = \frac{2 \to a}{5} - \to d$
$= \frac{1}{5} (2 \to a - 5 \to d)$