Question

The two vectors `hati + hatj + hatk` and `3hati - hatj + 3hatk` represent the two sides OA and OB, respectively, of a ∆OAB, where O is the origin. The point P lies on AB such that OP is a median. Find the area of the parallelogram formed by the two adjacent sides as OA and OP.

Answer 1

 

We need to find the area of the parallelogram formed by `vec(OP)` and `vec(OA)`.

Finding `vec(OP)`:

`vec(OP)` is the position vector of point P.

Since point P is the midpoint of AB.

Position vector of P = `(vec(OA) + vec(OB))/2`

`vec(OP) = ((hati + hatj + hatk) + (3hati - hatj + 3hatk))/2`

= `((1 + 3)hati + (1 - 1)hatj + (1 + 3) hatk)/2`

= `(4hati + 4hatk)/2`

= `(2(2hati + 2hatk))/2`

= `2hati + 2hatk`

Find the area of the parallelogram formed by `vec(OP)` and `vec(OA)`.

Area of the parallelogram = `|vec(OP) xx vec(OA)|`

Now, `vec(OP) xx vec(OA) = |(hati, hatj, hatk),(2, 0, 2),(1, 1, 1)|`

= `hati(0 xx 1 - 1 xx 2) - hatj(2 xx 1 - 1 xx 2) + hatk(2 xx 1 - 1 xx 0)`

= `-2hati - hatj xx 0 + 2hatk`

= `-2hati + 2hatk`

Magnitude of `vec(OP) xx vec(OA) = sqrt((-2)^2 + (2)^2)`

`|vec(OP) xx vec(OA)| = sqrt(4 + 4)`

= `sqrt8`

= `2sqrt2`

Area of the parallelogram = `|vec(OP) xx vec(OA)|`

= `2sqrt2` sq. units