The area of the triangle formed by the point 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 A $\sqrt{305}$ sq. units
$\to O A = - 3^i +^j$ , $\to O B = 5^i + 2^j +^k$ , $\to O C =^i - 2^j + 3^k$

$\to A B = \to O B - \to O A = 5^i + 2^j +^k + 3^i -^j = 8^i +^j +^k$
$\to B C = \to O C - \to O B =^i - 2^j + 3^k - 5^i - 2^j -^k = - 4^i - 4^j + 2^k$

$| \to A B \times \to B C | = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 8 & 1 & 1 - 4 & - 4 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$=^i (2 + 4) -^j (16 + 4) +^k (- 32 + 4)$
$= 6^i - 20^j - 28^k$
$| \to A B \times \to B C | = \sqrt{36 + 400 + 784} = \sqrt{1220}$
$= 2 \sqrt{305}$

Area of $Δ A B C = \frac{1}{2} | \to A B \times \to B C | = \frac{1}{2} \times 2 \sqrt{305}$

$= \sqrt{305}$ 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

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

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

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