Maths · Scalar and vector products

Consider with , B \equiv(\vec{b}) \) and , \) ff =\vec{b} \cdot \vec{b}+ \) then

Consider \( \Delta A B C \) with \( A \equiv(\vec{a}), B \equiv(\vec{b}) \) and \( C=(\vec{c}), \) ff \( \vec{b} .(\vec{a}+\vec{c})=\vec{b} \cdot \vec{b}+ \) \( \vec{a} \cdot \vec{c} ;|\vec{b}-\vec{a}|=3 ;|\vec{c}-\vec{b}|=4 \) then the angle between the medians \( \overline{A M} \) and \( B D \) is

  • A. \( \pi-\cos ^{-1}\left(\frac{1}{5 \sqrt{13}}\right) \)
  • B. \( \pi-\cos ^{-1}\left(\frac{1}{13 \sqrt{5}}\right) \)
  • C. \( \cos ^{-1}\left(\frac{1}{5 \sqrt{13}}\right) \)
  • D. \( \cos ^{-1}\left(\frac{1}{13 \sqrt{5}}\right) \)

Step-by-step solution

Given conditions imply triangle ABC is a 3-4-5 right triangle with right angle at B. Using coordinates B=(0,0), A=(3,0), C=(0,4), medians AM and BD have vectors (-3,2) and (1.5,2). Dot product = -0.5, magnitudes √13 and 2.5, giving cosθ = -1/(5√13). Thus θ = π - cos⁻¹(1/(5√13)).
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