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Applications of Cross Product

If i - j + ...

Question

If $i - j + 2 k, 2 i + j - k$ and $3 i - j + 2 k$ are position vectors of vertices of a triangle ,then its area is

26

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13

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2 \sqrt{13}

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\sqrt{13}

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Solution

The correct option is A $\sqrt{13}$
Given
$- - \to O A = ˆ i - ˆ j + 2 ˆ k$
$- - \to O B = 2 ˆ i + ˆ j - ˆ k$
$- - \to O C = 3 ˆ i - ˆ j + 2 ˆ k$
Now,
Area of triangle $= \frac{1}{2} ∣ ∣ ∣ - - \to A B \times - - \to B C ∣ ∣ ∣$

$- - \to A B = - - \to O B - - - \to O A$
$= (2 - 1) ˆ i + (1 + 1) ˆ j + (- 1 - 2) ˆ k$
$= ˆ i + 2 ˆ j - 3 ˆ k$
And,
$- - \to B C = - - \to O C - - - \to O B$
$= (3 - 2) ˆ i + (- 1 - 1) ˆ j + (2 + 1) ˆ k$
$= ˆ i - 2 ˆ j + 3 ˆ k$

Now
$- - \to A B \times - - \to B C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 1 & 2 & - 3 1 & - 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ˆ i (6 - 6) - ˆ j (3 + 3) + ˆ k (- 2 - 2)$
$= 0 - 6 ˆ j - 4 ˆ k$
So,
$∣ ∣ ∣ - - \to A B \times - - \to B C ∣ ∣ ∣ = \sqrt{{(6)}^{2} + {(4)}^{2}} = \sqrt{52}$
Hence, area of triangle is $\frac{1}{2}$ $\sqrt{52}$ = $\sqrt{13}$
Hence, the option $D$ is the correct answer.

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