If $\to a = i + 2 j + 2 k, | \to b | = 5$ and the angle between $\to a$ and $\to b$ is $\frac{π}{6}$ , then the area of the triangle formed by these two vectors as two sides is

$\frac{15}{2}$

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$15$

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$\frac{15}{4}$

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$\frac{15 \sqrt{3}}{2}$

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Solution

The correct option is C
$\frac{15}{4}$

$a r e a = \frac{1}{2} | a \times b | = \frac{1}{2} | a | | b | s i n \frac{π}{6} = \frac{15}{4}$

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Similar questions

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If $\to a = i + 2 j + 2 k, | \to b | = 5$ and the angle between $\to a$ and $\to b$ is $\frac{π}{6}$ , then the area of the triangle formed by these two vectors as two sides is

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If →a=i+2j+2k,|→b|=5 and the angle between →a and →b is π6, then the area of the triangle formed by these two vectors as two sides is

The correct option is C 154 area=12|a×b|=12|a||b|sinπ6=154

If $\to a = i + 2 j + 2 k, | \to b | = 5$ and the angle between $\to a$ and $\to b$ is $\frac{π}{6}$ , then the area of the triangle formed by these two vectors as two sides is

The correct option is C
$\frac{15}{4}$

$a r e a = \frac{1}{2} | a \times b | = \frac{1}{2} | a | | b | s i n \frac{π}{6} = \frac{15}{4}$