Find the area of the triangle whose adjacent sides are determined by the vector $\to a = - 2^i - 5^k$ and $\to b =^i - 2^j -^k$

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Solution

As we know that area of triangle when two consecutive sides $\to a = - 2 ˆ i - 5 ˆ k$ and $\to b = ˆ i - 2 ˆ j - ˆ k$ are given-

Area of $△ = \frac{1}{2} | \to a \times \to b |$
Area of $△ = \frac{1}{2} | (- 2 ˆ i - 5 ˆ k) \times (ˆ i - 2 ˆ j - ˆ k) |$

Area of $△ = \frac{1}{2} | [0 \times (- 2) - (- 5) (- 2)] ˆ i - [(- 1) (- 2) - (1) (- 5)] ˆ j + [(- 2) (- 2) - 0 \times 1] ˆ k |$

Area of $△ = \frac{1}{2} | [- 10] ˆ i - [+ 7] ˆ j + [+ 4] ˆ k |$

Area of $△ = \frac{1}{2} \sqrt{(- 10)^{2} - (+ 7)^{2} + (+ 4)^{2}}$

Area of $△ = \frac{1}{2} \sqrt{67} u n i t^{2}$

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Similar questions

Find the area of the parallelogram whose adjacent sides are determined by the vector .

\to a = 3^i +^j + 4^k

\to b =^i -^j +^k

\to a =^i - 2^j +^k, \to b = - 2^i + 4^j + 5^k

\to c =^i - 6^j - 7^k

\to r

\to a =^i - 2^j + 5^k

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

\to r . (2^i +^j +^k) + 8 = 0

Find the area of the parallelogram whose adjacent sides are determined by the vectors $a =^i -^j + 3^k$ and $b = 2^i - 7^j +^k$

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