# Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. - Physics

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Numerical

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

#### 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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