Test Index

CBSE Class 12 Math 2025 All Sets Solved Paper

© examsnet.com
Question : 9 of 20
Marks: +1, -0
If a⃗+b⃗+c⃗=0⃗,∣a⃗∣=37,∣b⃗∣=3\vec{a}+\vec{b}+\vec{c}=\vec{0}, \left|\vec{a}\right| = \sqrt{37}, \left|\vec{b}\right| = 3 and ∣c⃗∣=4,\left|\vec{c}\right| = 4, then angle between b⃗\vec{b} and c⃗\vec{c} is
Solution:  
Given, a⃗+b⃗+c⃗=0⃗,∣a⃗∣=37,∣b⃗∣=3,∣c⃗∣=4\text{Given, } \vec{a}+\vec{b}+\vec{c}=\vec{0}, \left|\vec{a}\right| = \sqrt{37}, \left|\vec{b}\right| = 3, \left|\vec{c}\right| = 4
b⃗+c⃗=−a⃗\vec{b}+\vec{c}=-\vec{a}
Squaring both sides,
∣b⃗∣2+∣c⃗∣2+2b⃗⋅c⃗=∣−a∣2\left|\vec{b}\right| ^{2}+ \left|\vec{c}\right| ^{2}+2 \vec{b}\cdot \vec{c}= \left| -a \right| ^{2}
⇒∣b⃗∣2+∣c⃗∣2+2∣b⃗∣∣c⃗∣cos⁡θ=∣a⃗∣2\Rightarrow \left|\vec{b}\right| ^{2}+ \left|\vec{c}\right| ^{2}+2 \left|\vec{b}\right|\left|\vec{c}\right| \cos \theta = \left|\vec{a}\right| ^{2}
⇒9+16+2(3)(4)cos⁡θ=37\Rightarrow 9+16+2(3)(4) \cos \theta =37
⇒24cos⁡θ=37−25\Rightarrow 24 \cos \theta =37-25
⇒24cos⁡θ=12\Rightarrow 24 \cos \theta =12
⇒    cos⁡θ=  12\Rightarrow \;\; \cos \theta =\;\frac{1}{2}
⇒θ=  π3\Rightarrow \theta =\;\frac{\pi}{3}
© examsnet.com
Go to Question:
Prev Question
Next Question