KCET · Maths · Vector Algebra
The vectors \(\mathbf{A B}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}\) and \(\mathbf{A C}=5 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) are the sides of a \(\triangle A B C\), The length of the median through \(A\) is
- A \(\sqrt{18}\)
- B \(\sqrt{72}\)
- C \(\sqrt{33}\)
- D \(\sqrt{288}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{33}\)
Step-by-step Solution
Detailed explanation
Let \(\mathbf{A}=\) origin
\(\therefore \mathbf{A B}=\) Position vector of \(\mathbf{B}\)
\(\mathbf{A C}=\) Position vector of \(\mathbf{C}\)
\(\therefore\) Position vector of mid-point of \(\mathbf{B}\) and \(\mathbf{C}\)
\(=\mathbf{P}=\frac{\mathbf{A B}+\mathbf{A C}}{2}\)
\(=\left(\frac{3+5}{2}\right) \hat{\mathbf{i}}+\left(\frac{0-2}{2}\right) \hat{\mathbf{j}}+\left(\frac{4+4}{2}\right) \hat{\mathbf{k}}\)
\(=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
Median \(=\mathbf{A P}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
Hence, length of median \(=|\mathbf{A P}|=\sqrt{16+1+16}\)
\(=\sqrt{33}\)
\(\therefore \mathbf{A B}=\) Position vector of \(\mathbf{B}\)
\(\mathbf{A C}=\) Position vector of \(\mathbf{C}\)
\(\therefore\) Position vector of mid-point of \(\mathbf{B}\) and \(\mathbf{C}\)
\(=\mathbf{P}=\frac{\mathbf{A B}+\mathbf{A C}}{2}\)
\(=\left(\frac{3+5}{2}\right) \hat{\mathbf{i}}+\left(\frac{0-2}{2}\right) \hat{\mathbf{j}}+\left(\frac{4+4}{2}\right) \hat{\mathbf{k}}\)
\(=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
Median \(=\mathbf{A P}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
Hence, length of median \(=|\mathbf{A P}|=\sqrt{16+1+16}\)
\(=\sqrt{33}\)
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