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Show that 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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#### Solution

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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