The area of the triangle formed by the points whose position vectors are $- 3^i +^j, 5^i + 2^j +^k$ and $^i - 2^j + 3^k$ is

\sqrt{23}

sq. units

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\sqrt{21}

sq. units

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\sqrt{305}

sq. units

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\sqrt{33}

sq. units

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Solution

The correct option is C $\sqrt{305}$ sq. units
$A = - 3 i + j$
$B = 5 i + 2 j + k$
$C = i - 2 j + 3 k$

Therefore
$\to A B = 8 i + j + k$
$\to B C = - 4 i - 4 j + 2 k$

Therefore area is
$= \frac{1}{2} | \to A B \times \to B C |$

$= \frac{1}{2} | 6 i - 20 j - 28 k |$

$= | 3 i - 10 j - 14 k |$

$= \sqrt{9 + 100 + 196}$

$= \sqrt{305}$ sq.units

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Similar questions

- 3^i +^j, 5^i + 2^j +^k

^i - 2^j + 3^k

\to A

^i + 2^j + 3^k

\to B

2^i - 3^j +^k

\to A = 3^i + 4^j

\to B = - 3^i + 7^j

The area of parallelogram whose adjacent sides are $^i +^k a n d 2^i +^j +^k$ is

Calculate the area of the triangle determined by the two vectors $\to A = 3^i + 4^j$ and $\to B = - 3^i + 7^j$

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