MHT CET · Maths · Vector Algebra
If the vectors \(\overline{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}\) and \(\overline{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) are the sides of the triangle ABC , then the length of the median, through \(A\), is
- A \(\sqrt{45}\) units.
- B \(\sqrt{18}\) units.
- C \(\sqrt{72}\) units.
- D \(\sqrt{33}\) units
Answer & Solution
Correct Answer
(D) \(\sqrt{33}\) units
Step-by-step Solution
Detailed explanation
Let \(A D\) be the median of \(\triangle A B C\).
\(\therefore \overline{\mathrm{AD}} =\frac{\overline{\mathrm{AB}}+\overline{\mathrm{AC}}}{2} \)
\( =\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}+5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{2} \)
\( =\frac{8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{2} \)
\( =4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \)
\( \therefore |\overline{\mathrm{AD}}| =\sqrt{4^2+(-1)^2+4^2}=\sqrt{33} \text { units}\)
\(\therefore \overline{\mathrm{AD}} =\frac{\overline{\mathrm{AB}}+\overline{\mathrm{AC}}}{2} \)
\( =\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}+5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{2} \)
\( =\frac{8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{2} \)
\( =4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \)
\( \therefore |\overline{\mathrm{AD}}| =\sqrt{4^2+(-1)^2+4^2}=\sqrt{33} \text { units}\)
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