If a- - - - 2 - - 2 k- - -b- - 5 and the angle between a- and b- is -6- then the area of the triangle formed by these two vectors as two sides is

The correct option is A 154 Area of the triangle =12 |a × b| = 12 ||a| |b| sinθn| = 12 [ 3 × 5 ×sinπ6 ] #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

If a⃗ = î +...

Question

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}{4}

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\frac{15}{2}

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15

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

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15 \sqrt{3}

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Solution

The correct option is A $\frac{15}{4}$
Area of the triangle
$= \frac{1}{2} | a \times b |$
$= \frac{1}{2} | | a | | b | sin θ ˆ n |$
$= \frac{1}{2} [3 \times 5 \times sin \frac{π}{6}]$
$(∴ | a | = \sqrt{1 + 2^{2} + 2^{2}} = 3)$
$= \frac{1}{2} [15 \times \frac{1}{2}] = \frac{15}{4}$

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Similar questions

Q. 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

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

Q. If

\to a =^i + 2^j - 3^k

and

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then the angle between the vectors

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is.

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and

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. The angles of the triangle are
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{cos}^{- 1} \frac{7}{\sqrt{75}}

{cos}^{- 1} (\sqrt{\frac{26}{75}})

{cos}^{- 1} \frac{3}{\sqrt{15}}

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If →a=^i+2^j+2^k,|→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 A 154Area of the triangle=12|a×b|=12||a||b|sinθˆn|=12[3×5×sinπ6] (∴|a|=√1+22+22=3)=12[15×12]=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 A $\frac{15}{4}$
Area of the triangle
$= \frac{1}{2} | a \times b |$
$= \frac{1}{2} | | a | | b | sin θ ˆ n |$
$= \frac{1}{2} [3 \times 5 \times sin \frac{π}{6}]$
$(∴ | a | = \sqrt{1 + 2^{2} + 2^{2}} = 3)$
$= \frac{1}{2} [15 \times \frac{1}{2}] = \frac{15}{4}$