Name: Types of Vectors
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: A vector is a quantity that has both magnitude and direction. Did you know that there are several types of vectors? Watch the video to learn about them.

Question

In a quadrilateral

A B C D

, if the midpoints of one pair of opposite sides and points of intersection of diagonals are collinear, using vector methods prove that the quadrilateral

A B C D

is a trapezium.

Accepted Answer

Let $ABCD$ be a quadrilateral. $A B C D$ are the midpoints of the sides $BC, AD$ respectively. $R$ be the point of intersection of the diagonals. Given that $M, R, N$ are collinear.

Let $A$ be the origin

$\bar{AB} = \bar{b}, \bar{AC} = \bar{c} and \bar{AD} = \bar{d}$

$\bar{AM} = \frac{\bar{b} + \bar{c}}{2}, \bar{AN} = \frac{\bar{d}}{2}$

Also $\bar{AR} = s, \bar{AC} = s \bar{c}$ where $s$ is scalar

Since $M, R, N$ are collinear

$\bar{MN} = t (\bar{RN})$ where $t$ is a scalar

$\bar{AN} - \bar{AM} = t (\bar{AN} - \bar{AR})$

$\frac{\bar{d}}{2} - \frac{\bar{b} + \bar{c}}{2} = t (\frac{\bar{d}}{2} - s \bar{c})$

$\frac{\bar{d} - \bar{b} - \bar{c}}{2} = \frac{t}{2} [\bar{d} - 2 s \bar{c}]$

$\bar{d} - \bar{b} - \bar{c} = t (\bar{d} - 2 s \bar{c})$

$\bar{d} - \bar{b} - \bar{c} = t \bar{d} - 2 st \bar{c}$

$(1 - t) \bar{d} + (- 1 + 2 st) \bar{c} = \bar{b} . . . (1)$

Let $R$ be divides $BD$ in the ratio $k : 1$

$\bar{AR} = \frac{k \bar{d} + \bar{b}}{k + 1} but \bar{AR} = s \bar{c}$

$\Rightarrow \frac{k \bar{d} + \bar{b}}{k + 1} = s \bar{c}$

$\Rightarrow k \bar{d} + \bar{b} = s (k + 1) \bar{c}$

$\Rightarrow \bar{b} = s (k + 1) \bar{c} - k \bar{d} . . . . . . (2)$

$(1) = (2)$

$\Rightarrow (1 - t) \bar{d} + (2 st - 1) \bar{c} = s (k + 1) \bar{c} - k \bar{d}$

$2 st - 1 = s (k + 1) . . . (i)$

$1 - t = - k \Rightarrow k + 1 = t$

Put $k + 1 = t in (i)$

$\Rightarrow 2 st - 1 = st \Rightarrow st = 1$

Substituting $st = 1$ in equation $(1)$

$\Rightarrow (1 - t) \bar{d} + \bar{c} = \bar{b}$

$\Rightarrow (1 - t) \bar{d} = \bar{b} - \bar{c}$

$\Rightarrow - (t - 1) \bar{d} = - (\bar{c} - \bar{b})$

$\Rightarrow (t - 1) \bar{AD} = \bar{AC} - \bar{AB} = \bar{BC}$

$\Rightarrow \bar{BC} = (t - 1) \bar{AD}$

$\Rightarrow \bar{A D} ∥ \bar{B C}$

$\Rightarrow ABCD$ is a trapezium.

In a quadrilateral $A B C D$ , if the midpoints of one pair of opposite sides and points of intersection of diagonals are collinear, using vector methods prove that the quadrilateral $A B C D$ is a trapezium.

Important Questions on Addition of Vectors

Learn and Prepare for any exam you want !

In a quadrilateral ABCD, if the midpoints of one pair of opposite sides and points of intersection of diagonals are collinear, using vector methods prove that the quadrilateral ABCD is a trapezium.

Important Questions on Addition of Vectors

Learn and Prepare for any exam you want !

In a quadrilateral $A B C D$ , if the midpoints of one pair of opposite sides and points of intersection of diagonals are collinear, using vector methods prove that the quadrilateral $A B C D$ is a trapezium.