AP EAMCET · Maths · Three Dimensional Geometry
\(3 \hat{i}-2 \hat{j}-\hat{k},-2 \hat{i}-\hat{j}+3 \hat{k}\) and \(-\hat{i}+3 \hat{j}-2 \hat{k}\) are the position vectors of the vertices \(A, B\) and \(C\) of a \(\triangle A B C\) respectively.
If \(\mathrm{H}\) is its orthocenter, then \(\overrightarrow{\mathrm{HA}}+\overrightarrow{\mathrm{HB}}+\overrightarrow{\mathrm{HC}}=\)
- A \(2 \overrightarrow{\mathrm{SA}}\)
- B \(\overrightarrow{0}\)
- C \(2 \overrightarrow{\mathrm{AB}}\)
- D \(\hat{i}+\hat{j}+\hat{k}\)
Answer & Solution
Correct Answer
(B) \(\overrightarrow{0}\)
Step-by-step Solution
Detailed explanation
Let \(H(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})\) be the orthocentre of the \(\triangle \mathrm{ABC}\). Since \(\overrightarrow{A H} \cdot \overrightarrow{B C}=0\) \(\{\because \overrightarrow{A H} \perp \overrightarrow{B C}\}\)…
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