MHT CET · Maths · Three Dimensional Geometry
The vectors \(\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}}+4 \hat{\mathrm{k}}\) are the sies of a triangle \(\mathrm{ABC}\). The length of the median through \(\mathrm{A}\) is
- A \(\sqrt{33}\) unit
- B \(\sqrt{288}\) unit
- C \(\sqrt{18}\) unit
- D \(\sqrt{72}\) unit
Answer & Solution
Correct Answer
(A) \(\sqrt{33}\) unit
Step-by-step Solution
Detailed explanation
Let \(\mathrm{A}\) be the origin.
Then \(\mathrm{B}=(3,0,4)\) and \(\mathrm{C}=(5,-1,4)\)
Mid point of \(\mathrm{BC}=\left(\frac{3+5}{2}, \frac{0-2}{2}, \frac{4+4}{2}\right)\) i.e. \((4,-1,4)\)
\(\therefore\) Length of medium \(=\sqrt{(4)^2+(-1)^2+(4)^2}=\sqrt{33}\)
Then \(\mathrm{B}=(3,0,4)\) and \(\mathrm{C}=(5,-1,4)\)
Mid point of \(\mathrm{BC}=\left(\frac{3+5}{2}, \frac{0-2}{2}, \frac{4+4}{2}\right)\) i.e. \((4,-1,4)\)
\(\therefore\) Length of medium \(=\sqrt{(4)^2+(-1)^2+(4)^2}=\sqrt{33}\)
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