JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(A(3, 0, -1), B(2, 10, 6)\) and \(C(1, 2, 1)\) be the vertices of a triangle and \(M\) be the midpoint of \(AC\). If \(G\) divides \(BM\) in the ratio, \(2 : 1\), then \(\cos \,\left( {\angle GOA} \right)\) (\(O\) being he origin) is equal to
- A \(\frac{1}{{\sqrt {30} }}\)
- B \(\frac{1}{{2\sqrt {15} }}\)
- C \(\frac{1}{{6\sqrt {10} }}\)
- D \(\frac{1}{{\sqrt {15} }}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{{\sqrt {15} }}\)
Step-by-step Solution
Detailed explanation
\(G\) will be centroid of \(\triangle \mathrm{ABC}\) \(G \equiv(2,4,2)\) \(\overrightarrow{\mathrm{OG}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{OA}}=3 \mathrm{i}-\hat{\mathrm{k}}\)…
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