If \(\vec A+3\vec B\) and \(4\vec A-\vec B\) are perpendicular and |A|

1.	If \(\vec A+3\vec B\) and \(4\vec A-\vec B\) are perpendicular and \|A\|=2\|B\|. Determine the angle between the non-Zero vectors \(\vec A\,and\,\vec B\)
Answer» Given \(\vec{A}+3\vec{B}\) and \(4\vec{A}-\vec{B}\) are perpendicular then, \((\vec{A}+3\vec{B})\).\((4\vec{A}-\vec{B})\) = 0 (\(\vec{a}.\vec{b}\)= ab cosθ if vectors are perpendicular then \(\vec{a}.\vec{b}\)= 0) \(\Rightarrow\) 4A² + 12\(\vec{B}.\vec{A}\)- \(\vec{A}.\vec{B}\) - \(3B^2\) = 0 (\(\because\) \(\vec{A}.\vec{A}=(\vec{A})^2=A^2\)) \(\Rightarrow\) 4A^{2 +}11\(\vec{A}.\vec{B}\) - \(3B^2\) = 0 (\(\because\) \(\vec{A}.\vec{B}\) = \(\vec{B}.\vec{A}\)) \(\Rightarrow\) 4A² + 11 AB cosθ - 3B² = 0 , where θ is angle between vectors \(\vec{A}\) and \(\vec{B}\) and \(\vec{A}.\vec{B}\) = AB cosθ \(\Rightarrow\) 16 B² + 22 B² cosθ - 3B² = 0 ( \(\because\) \(\|\vec{A}\|=2\|\vec{B}\|\) = A = 2B) \(\Rightarrow\) B²(13 + 22 cosθ) = 0 \(\Rightarrow\) 13 + 22 cosθ = 0 ( \(\because\) \(\vec{B}\) is non- zero vector therefore B\(\ne\)0) \(\Rightarrow\) cosθ = \(-\frac{13}{22}\) \(\Rightarrow\) θ = \(\cos^{-1}\big(-\frac{13}{22}\big)\) = π - \(\cos^{-1}\big(\frac{13}{22}\big)\) (\(\because\) cos^-1(-θ) = π - cos^-1θ)

If \(\vec A+3\vec B\) and \(4\vec A-\vec B\) are perpendicular and |A|=2|B|. Determine the angle between the non-Zero vectors \(\vec A\,and\,\vec B\)

Answer»

Given \(\vec{A}+3\vec{B}\) and \(4\vec{A}-\vec{B}\) are perpendicular

then, \((\vec{A}+3\vec{B})\).\((4\vec{A}-\vec{B})\) = 0 (\(\vec{a}.\vec{b}\)= ab cosθ if vectors are perpendicular then \(\vec{a}.\vec{b}\)= 0)

\(\Rightarrow\) 4A² + 12\(\vec{B}.\vec{A}\)- \(\vec{A}.\vec{B}\) - \(3B^2\) = 0 (\(\because\) \(\vec{A}.\vec{A}=(\vec{A})^2=A^2\))

\(\Rightarrow\) 4A^{2 +}11\(\vec{A}.\vec{B}\) - \(3B^2\) = 0 (\(\because\) \(\vec{A}.\vec{B}\) = \(\vec{B}.\vec{A}\))

\(\Rightarrow\) 4A² + 11 AB cosθ - 3B² = 0 , where θ is angle between vectors \(\vec{A}\) and \(\vec{B}\) and \(\vec{A}.\vec{B}\) = AB cosθ

\(\Rightarrow\) 16 B² + 22 B² cosθ - 3B² = 0 ( \(\because\) \(|\vec{A}|=2|\vec{B}|\) = A = 2B)

\(\Rightarrow\) B²(13 + 22 cosθ) = 0

\(\Rightarrow\) 13 + 22 cosθ = 0 ( \(\because\) \(\vec{B}\) is non- zero vector therefore B\(\ne\)0)

\(\Rightarrow\) cosθ = \(-\frac{13}{22}\)

\(\Rightarrow\) θ = \(\cos^{-1}\big(-\frac{13}{22}\big)\) = π - \(\cos^{-1}\big(\frac{13}{22}\big)\) (\(\because\) cos^-1(-θ) = π - cos^-1θ)