KCET · Maths · Area Under Curves
The area of the parallelogram whose adjacent sides are \( \hat{i}+\hat{k} \) and \( 2 \hat{i}+\hat{j}+\hat{k} \) is
- A ( \( \sqrt{2} \)
- B \( \sqrt{3} \)
- C \( 13 \)
- D \( 04 \)
Answer & Solution
Correct Answer
(B) \( \sqrt{3} \)
Step-by-step Solution
Detailed explanation
Given sides of parallelogram,
\[
\begin{array}{l}
\vec{a}=\hat{i}+\hat{k} \\
\vec{b}=2 \hat{i}+\hat{j}+\hat{k}
\end{array}
\]
So, area is given by \( |\vec{a} \times \vec{b}| \)
\[
\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 1 \\
2 & 1 & 1
\end{array}\right|=-\hat{i}+\hat{j}+\hat{k}
\]
Therefore, area of parallelogram, is \( \sqrt{1+1+1}=\sqrt{3} \)
\[
\begin{array}{l}
\vec{a}=\hat{i}+\hat{k} \\
\vec{b}=2 \hat{i}+\hat{j}+\hat{k}
\end{array}
\]
So, area is given by \( |\vec{a} \times \vec{b}| \)
\[
\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 1 \\
2 & 1 & 1
\end{array}\right|=-\hat{i}+\hat{j}+\hat{k}
\]
Therefore, area of parallelogram, is \( \sqrt{1+1+1}=\sqrt{3} \)
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