Name: Geometrical Interpretation of Scalar Triple Product
Uploaded: 2019-07-19
Duration: 14 min
Description: Do you know what is the geometrical meaning of the scalar triple product? And why the scalar triple product so important? Check out the video to explore.

Question

Prove that the straight line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and half of their difference.

Accepted Answer

Consider a trapezium $P Q R S$ . Let $P R$ and $Q S$ be the two diagonals. Let $T$ and $U$ be the mid-points of diagonals $P R$ and $Q S$ respectively. Consider the position vectors of the points $P, Q, R$ and $S$ as $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ , respectively. Draw a figure representing all these essential details.

The position vectors of $T$ and $U$ can be found as

$\vec{T} = \frac{\vec{a} + \vec{c}}{2}$ and $\vec{U} = \frac{\vec{b} + \vec{d}}{2}$

We have to prove that $T U$ is parallel to $P S$ and $Q R$ . The vectors $P S$ and $Q R$ can be computed as

$\vec{P S} = \vec{d} - \vec{a} \dots \dots \dots \dots \dots (1)$

$\vec{Q R} = \vec{c} - \vec{b} \dots \dots \dots \dots \dots (2)$

Now vector $T U$ is obtained as

$\vec{T U} = \frac{\vec{b} + \vec{d}}{2} - \frac{\vec{a} + \vec{c}}{2}$

$= \frac{1}{2} [\vec{b} + \vec{d} - \vec{a} - \vec{c}]$

$= \frac{1}{2} [(\vec{d} - \vec{a}) + (\vec{b} - \vec{c})]$

Substituting the vectors $P S$ and $Q R$ from equations $1$ and $2$ in the above equation,

$\vec{T U} = \frac{1}{2} [\vec{P S} - \vec{Q R}]$

$= \frac{1}{2} \vec{P S} - \frac{1}{2} \vec{Q R}$

As $\vec{P S}$ is parallel to $\vec{Q R}$ . $\vec{P S} = λ \vec{Q R}, λ > 0, λ \neq 1$ .

$\vec{T U} = \frac{\vec{Q R}}{2} [λ - 1]$

Hence, they are parallel to each other.

Also, the equation clearly shows that the line segment joining the mid-points of the diagonals is equal to half the difference of the parallel sides.

Prove that the straight line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and half of their difference.

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