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प्रश्न
Show that area of the parallelogram whose diagonals are given by `vec"a"` and `vec"b"` is `(|vec"a" xx vec"b"|)/2`. Also find the area of the parallelogram whose diagonals are `2hat"i" - hat"j" + hat"k"` and `hat"i" + 3hat"j" - hat"k"`.
उत्तर
Let ABCD be a parallelogram such that,
`vec"AB" = vec"p"`
`vec"AD" = vec"q" = vec"BC"`
∴ By law of triangle, we get
`vec"AC" = vec"a" = vec"p" + vec"q"` .....(i)
And `vec"BD" = vec"b" = -vec"p" + vec"q"` .....(ii)
Adding equation (i) and (ii) we get,
`vec"a" + vec"b" = 2vec"q"`
⇒ `vec"q" = ((vec"a" + vec"b")/2)`
Subtracting equation (ii) from equation (i) we get
`vec"a" - vec"b" = 2vec"p"`
⇒ `vec"p" = ((vec"a" - vec"b")/2)`
∴ `vec"p" xx vec"q" = 1/4(vec"a" + vec"b") xx (vec"a" - vec"b")`
= `1/4 (vec"a" xx vec"a" - vec"a" xx vec"b" + vec"b" xx vec"a" - vec"b" xx vec"b")`
= `1/4(-vec"a" xx vec"b" xx vec"b" xx vec"a")` ......`[(because vec"a" xx vec"a" = 0),(vec"b" xx vec"b" = 0)]`
= `1/4(vec"a" xx vec"b" + vec"a" xx vec"b")`
= `1/4 * 2(vec"a" xx vec"b")`
= `|vec"a" xx vec"b"|/2`
So, the area of the parallelogram ABCD = `|vec"p" xx vec"q"| = 1/2|vec"a" xx vec"b"|`
Now area of parallelogram whose diagonals are `2hat"i" - hat"j" + hat"k"` and `hat"i" + 3hat"j" - hat"k"`
= `1/2|(2hat"i" - hat"j" + hat"k") xx (hat"i" + 3hat"j" - hat"k")|`
= `-|(hat"i", hat"j", hat"k"),(2, 1, 1),(1, 3, 1)|`
= `1/2 |hat"i"(1 - 3) - hat"j"(-2 - 1) + hat"k"(6 + 1)|`
= `1/2 - 2hat"i" + 3hat"j" + 7hat"k"|`
= `1/2 sqrt((-2)^2 + (3)^2 + (7)^2)`
= `1/2 sqrt(4 + 9 + 49)`
= `1/2 sqrt(62)` sq.units
Hence, the required area is `1/2 sqrt(62)` sq.units
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