Calculate the area of the triangle determined by the two vectors $\to A =^i - 2^j +^k$ and $\to B = 2^i + 3^j -^k$

^i + 3^j - 7^k

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\frac{1}{2} (-^i - 3^j + 7^k)

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\frac{1}{2} (-^i + 3^j + 7^k)

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-^i + 3^j + 7^k

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Solution

The correct option is C $\frac{1}{2} (-^i + 3^j + 7^k)$
We know that the area of the triangle is given by $\frac{1}{2} (\to A \times \to B)$
Thus, we have $\to A \times \to B$ as
$⎡ ⎢ ⎣ \begin{matrix} ^i & ^j & ^k 1 & - 2 & 1 2 & 3 & - 1 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow (2 - 3)^i - (- 1 - 2)^j + (3 - (- 4))^k$
$\Rightarrow -^i + 3^j + 7^k$
Hence, the area of the triangle is
$\frac{1}{2} (\to A \times \to B) = \frac{1}{2} (-^i + 3^j + 7^k)$

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Similar questions

\to A =^i - 2^j +^k

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

The given vectors can form sides of a triangle $^i -^j -^k, - 2^i + 3^j -^k,^i - 2^j + 2^k$

Calculate the area of the parallelogram when adjacent sides are given by the vectors $\to A$ = $^i + 2^j + 3^k$ and $\to B$ = $2^i - 3^j +^k$

The given vectors can form sides of a triangle $^i -^j -^k, - 2^i + 3^j -^k,^i - 2^j + 2^k$

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

\to b =^i +^k

\to c = 2^j -^k

A

(\to a + \to b)

(\to b + \to c)

4 A^{2}

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