In a trapezium, the vector BC = AD . We will then find that p = A… - Vidyadip

MCQ

In a trapezium, the vector $\overrightarrow {BC} = \lambda \overrightarrow {AD} .$ We will then find that $p = \overrightarrow {AC} + \overrightarrow {BD} $ is collinear with $\overrightarrow {AD} ,$ If $p = \mu \overrightarrow {AD} ,$ then

✓
$\mu = \lambda + 1$
B
$\lambda = \mu + 1$
C
$\lambda + \mu = 1$
D
$\mu = 2 + \lambda $

✓

Answer

Correct option: A.

$\mu = \lambda + 1$

a
(a) We have, $p = \overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {AC} + \overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {AC} + \lambda \overrightarrow {AD} + \overrightarrow {CD} $

$ = \lambda \,\overrightarrow {AD} + (\overrightarrow {AC} + \overrightarrow {CD} ) = \lambda \,\overrightarrow {AD} + \overrightarrow {AD} = (\lambda + 1)\overrightarrow {AD} .$

Therefore $p = \mu \overrightarrow {AD} \Rightarrow \mu = \lambda + 1.$

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