Question

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

• A. $\mu = \lambda + 1$
• B. $\lambda = \mu + 1$
• C. $\lambda + \mu = 1$
• D. $\mu = 2 + \lambda$

Answer 1

Answer: (A)

$\mu = \lambda + 1$

Answer 2

We have, $\mathbf{p} = \overset{\rightarrow}{AC} + \overset{\rightarrow}{BD} = \overset{\rightarrow}{AC} + \overset{\rightarrow}{BC} + \overset{\rightarrow}{CD} = \overset{\rightarrow}{AC} + \lambda\overset{\rightarrow}{AD} + \overset{\rightarrow}{CD}$

$= \lambda\overset{\rightarrow}{AD} + (\overset{\rightarrow}{AC} + \overset{\rightarrow}{CD}) = \lambda\overset{\rightarrow}{AD} + \overset{\rightarrow}{AD} = (\lambda + 1)\overset{\rightarrow}{AD}.$

Therefore $\mathbf{p} = \mu\overset{\rightarrow}{AD} \Rightarrow \mu = \lambda + 1.$