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Standard XII

Mathematics

Determinant Form of a Line

Show that the...

Question

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

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Solution

Consider two vectors OK = vector $| a |$ and OM = vector $| b |$ , inclined at an angle $θ$ as shown in the following figure.
In $△ O M N,$ we can write the relation:
$s i n θ = \frac{M N}{O M} = \frac{M N}{∣ ∣ \to b ∣ ∣}$
$⟹ M N = ∣ ∣ \to b ∣ ∣ s i n θ$

$∣ ∣ \to a \times \to b ∣ ∣ = | \to a | ∣ ∣ \to b ∣ ∣ s i n θ$
$= O K \times M N = 2 \times \frac{1}{2} \times O K \times M N$
$= 2 \times$ Area of $△ O M K$

$⟹$ Area of $△$ $O M K = \frac{1}{2} \times ∣ ∣ \to a \times \to b ∣ ∣$

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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.

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