MHT CET · Maths · Vector Algebra
If the vectors \(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}},-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}\) represents
the diagonals of a parallelogram, then its area will be
- A \(\sqrt{21}\)
- B \(\frac{\sqrt{21}}{2}\)
- C \(2 \sqrt{21}\)
- D \(\frac{\sqrt{21}}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{21}}{2}\)
Step-by-step Solution
Detailed explanation
Let
\(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}\)
Now, \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \mathbf{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0\end{array}\right|\)
\(=-4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
Hence, required area \(=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|\)
\(=\frac{1}{2} \sqrt{16+4+1}=\frac{\sqrt{21}}{2}\)
\(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}\)
Now, \(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \mathbf{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0\end{array}\right|\)
\(=-4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
Hence, required area \(=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|\)
\(=\frac{1}{2} \sqrt{16+4+1}=\frac{\sqrt{21}}{2}\)
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