Question

Let `bar a = hati + hatj + hatk, barb = hati + 3hatj + 5hatk and barc = 7hati + 9hatj + 11hatk` then the area of parallelogram having diagonals `bara + barb and barb + barc` is

Options

• `4sqrt6` sq units

• `4sqrt6` sq units

• `sqrt6` sq units

• `6sqrt6` sq units

Answer 1

`bb(4sqrt6)` sq units

Explanation:

Given `bar a = hati + hatj + hatk`

`barb = hati + 3hatj + 5hatk`

`barc = 7hati + 9hatj + 11hatk`

Then, `bara + barb = 2hati + 4hatj + 6hatk`

and `barb = barc = 8hati + 12hatj + 16hatk`

Area of parallelogram = `1/2 |(bara + barb) xx (barb + barc)|`

`1/2 |2(hati + 2hatj + 3hatk) xx 4(2hati + 3hatj + 4hatk)|`

= `8/2 |(hati + 2hatj + 3hatk) xx (2hati + 3hatj + 4hatk)|`

= `4|(hati, hatj, hatk),(1, 2, 3),(2, 3, 4)| = 4|-hati + 2hatj - hatk|`

= `4sqrt(1 + 4 + 1)`

= `4sqrt6` sq. units.