AP EAMCET · Maths · Vector Algebra
If three consecutive vertices of a parallelogram are \(A(4,3,5), B(0,6,0)\), \(C(-8,1,4)\) and \(D\) is the fourth vertex, then the angle between \(\mathbf{A C}\) and \(\mathbf{B D}\) is
- A \(\cos ^{-1}\left(\frac{65}{\sqrt{149} \sqrt{161}}\right)\)
- B \(\cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}\right)\)
- C \(\cos ^{-1}\left(\frac{73}{\sqrt{149} \sqrt{161}}\right)\)
- D \(\cos ^{-1}\left(\frac{15}{\sqrt{149} \sqrt{161}}\right)\)
Answer & Solution
Correct Answer
(B) \(\cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}\right)\)
Step-by-step Solution
Detailed explanation
Given, \(A(4,3,5), B(0,6,0), C(-8,1,4)\) and \(D\) are the vertices of a parallelogram Let \(D\) be the point \((x, y, z)\). \(\therefore\) Diagonals of parallelogram bisect each other. \(\Rightarrow\) mid-point of \(A C=\) mid point of \(B D\)…
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