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प्रश्न
Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
उत्तर
Let `vec"OA" = vec"a" ; vec"OB" = vec"b" , angle"AOB" = theta`
As shown in the figure, ΔOAB is formed between the vectors `vec"a"` and `vec"b"` while OACB is a parallelogram.
From the definition,
`vec"a" xx vec"b" = |vec"a"| |vec"b"| sin thetahat"n"`
where `hat"n"` is the unit vector perpendicular to both vectors `vec"a"` and `vec"b"`
∴ `|vec"a" xx vec"b"| = |vec"a"| |vec"b"| sin theta`
= OA . OB sin θ ...(1)
In the figure, BD is perpendicular to OA.
∴ `"sin" theta = "BD"/"OB"` या `"BD" = "OB" "sin" theta`
From equation (1),
`|vec"a" xx vec"b"| = "OA" . "BD"`
= Base of parallelogram × Perpendicular distance between parallel sides
= Area of parallelogram OACB
But area of ΔOAB = `1/2` area of parallelogram OACB
= `1/2 |vec"a" xx vec"b"|`
Thus, the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
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