If \( \vec{a},\vec{b},\vec{c} \) are three vectors such that \( |\vec{a}|=2,\ |\vec{b}|=5,\ |\vec{c}|=3 \) and \( \vec{a}+2\vec{b}-3\vec{c}=\vec{0} \), then angle between \( \vec{b} \) and \( \vec{c} \) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \cos^{-1}\left(\displaystyle\frac{177}{180}\right) \)
Since \( \vec{a}+2\vec{b}-3\vec{c}=\vec{0} \), the vector \( \vec{a} \) can be written in terms of \( \vec{b} \) and \( \vec{c} \):
\( \displaystyle \vec{a}=3\vec{c}-2\vec{b} \)
Taking the dot product of each side with itself:
\( \displaystyle |\vec{a}|^{2}=9|\vec{c}|^{2}+4|\vec{b}|^{2}-12\,\vec{b}\cdot\vec{c} \)
\( \displaystyle 4=9(9)+4(25)-12\,\vec{b}\cdot\vec{c} \)
\( \displaystyle 4=181-12\,\vec{b}\cdot\vec{c} \)
\( \displaystyle \vec{b}\cdot\vec{c}=\frac{177}{12} \)
If θ is the angle between \( \vec{b} \) and \( \vec{c} \):
\( \displaystyle \cos\theta=\frac{\vec{b}\cdot\vec{c}}{|\vec{b}||\vec{c}|} \)
\( \displaystyle =\frac{177/12}{5\times 3} \)
\( \displaystyle =\frac{177}{180} \)
\( \displaystyle \theta=\cos^{-1}\left(\frac{177}{180}\right) \)
Why the other options are incorrect
Option (1) \( \cos^{-1}\left(\displaystyle\frac{-23}{40}\right) \) — Incorrect. This is the angle between \( \vec{a} \) and \( \vec{b} \), the wrong pair: from \( 3\vec{c}=\vec{a}+2\vec{b} \), \( 81=4+100+4\,\vec{a}\cdot\vec{b} \), so \( \vec{a}\cdot\vec{b}=-\frac{23}{4} \) and \( \cos\theta=\frac{-23/4}{10}=-\frac{23}{40} \).
Option (2) \( \cos^{-1}\left(-\displaystyle\frac{177}{180}\right) \) — Incorrect. \( \vec{b}\cdot\vec{c}=\frac{177}{12} \) is positive, so the cosine cannot be negative.
Option (3) \( \cos^{-1}\left(\displaystyle\frac{23}{40}\right) \) — Incorrect. The cosine of the angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{177}{180} \), not \( \frac{23}{40} \).