KCET · Maths · Vector Algebra
Suppose \( \vec{a}+\vec{b}+\vec{c}=0,|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=7 \), then the angle between \( \vec{a} \& \vec{b} \) is
- A \( \pi \)
- B \( \frac{\pi}{2} \)
- C \( \frac{\pi}{3} \)
- D \( \frac{\pi}{4} \)
Answer & Solution
Correct Answer
(C) \( \frac{\pi}{3} \)
Step-by-step Solution
Detailed explanation
Given that, \(\vec{a}+\vec{b}+\vec{c}=0\)
and \(|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=7\)
Now, \(\vec{a}+\vec{b}=-\vec{c}\)
So,
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(-\vec{c}) \cdot(-\vec{c})\)
\(\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{c}|^{2}\)
\(\Rightarrow 9+25+2(3)(5) \cos \theta=49\)
\(\Rightarrow \cos \theta=\frac{15}{2 \times 3 \times 5}=\frac{1}{2}\)
\(\Rightarrow \theta=\frac{\Pi}{3}\)
and \(|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=7\)
Now, \(\vec{a}+\vec{b}=-\vec{c}\)
So,
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=(-\vec{c}) \cdot(-\vec{c})\)
\(\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{c}|^{2}\)
\(\Rightarrow 9+25+2(3)(5) \cos \theta=49\)
\(\Rightarrow \cos \theta=\frac{15}{2 \times 3 \times 5}=\frac{1}{2}\)
\(\Rightarrow \theta=\frac{\Pi}{3}\)
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