MHT CET · Maths · Vector Algebra
If \(\overline{A B}=3 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}, \overline{A C}=5 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}\) represent the sides of triangle \(\mathrm{ABC}\), then the length of median through \(\mathrm{A}\) is
- A \(\sqrt{6}\) units
- B 5 units
- C \(\sqrt{5}\) units
- D 6 units
Answer & Solution
Correct Answer
(B) 5 units
Step-by-step Solution
Detailed explanation
Given
\(\begin{aligned}
\frac{\overline{\mathrm{AB}}}{\mathrm{AC}} &=3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\
&=5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}\)
Let \(\overline{\mathrm{AD}}\) is median
position vector of
\(\begin{aligned}
\overline{\mathrm{AD}} &=\frac{(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{2} \\
\overline{\mathrm{AD}} &=4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}} \\
\therefore|\overline{\mathrm{AD}}| &=\sqrt{16+9}=\sqrt{25}=5
\end{aligned}\)
\(\begin{aligned}
\frac{\overline{\mathrm{AB}}}{\mathrm{AC}} &=3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\
&=5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{aligned}\)
Let \(\overline{\mathrm{AD}}\) is median
position vector of
\(\begin{aligned}
\overline{\mathrm{AD}} &=\frac{(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{2} \\
\overline{\mathrm{AD}} &=4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}} \\
\therefore|\overline{\mathrm{AD}}| &=\sqrt{16+9}=\sqrt{25}=5
\end{aligned}\)
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