The vectors from origin to the points A and B are a = 2 i - 3 j + 2 k and b = 2 i + 3 j + k respectively, then the area of O A B is equal to:

Question

The vectors from origin to the points A and B are $\vec{a} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k}$ and $\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ respectively, then the area of $∆ O A B$ is equal to:

Smart Achievers · Accepted Answer

(4)

Ans. $\frac{1}{2} \sqrt{229}$

Explanation:

$∴$ $Area of ∆ O A B$

$= \frac{1}{2}$ $| \vec{O A} \times \vec{O B} |$

$= \frac{1}{2}$ $| (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \times (2 \hat{i} + 3 \hat{j} + \hat{k}) |$

$= \frac{1}{2}$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 2 \\ 2 & 3 & 1 \end{matrix} |$

$= \frac{1}{2}$ $| \hat{i} (- 3 - 6) - \hat{j} (2 - 4) + \hat{k} (6 + 6) |$

$= \frac{1}{2}$ $| - 9 \hat{i} + 2 \hat{j} + 12 \hat{k} |$

$∴$ $Area of ∆ O A B$

$= \frac{1}{2} \sqrt{81 + 4 + 144} = \frac{1}{2} \sqrt{229}$

Solution

Q.
The vectors from origin to the points A and B are $\vec{a} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k}$ and $\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ respectively, then the area of $∆ O A B$ is equal to:

1 $340$

2 $\sqrt{25}$

3 $\sqrt{229}$

4 $\frac{1}{2} \sqrt{229}$

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Solution

Q. The vectors from origin to the points A and B are a→=2i^-3j^+2k^ and b→=2i^+3j^+k^ respectively, then the area of ∆OAB is equal to:

1 340

2 25

3 229

4 12229