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If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three vectors such that ∣a⃗∣=2,∣b⃗∣=5,∣c⃗∣=3|\vec{a}| = 2, |\vec{b}| = 5, |\vec{c}| = 3 and a⃗+2b⃗−3c⃗=0⃗\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0}, then angle between b⃗\vec{b} and c⃗\vec{c} is equal to:

Solution

✅ Correct Option: 4

From the relation, a⃗=3c⃗−2b⃗\vec a = 3\vec c - 2\vec b, so ∣a⃗∣2=9∣c⃗∣2+4∣b⃗∣2−12(b⃗⋅c⃗)|\vec a|^2 = 9|\vec c|^2 + 4|\vec b|^2 - 12(\vec b\cdot\vec c).

Thus 4=81+100−12(b⃗⋅c⃗)4 = 81 + 100 - 12(\vec b\cdot\vec c), giving b⃗⋅c⃗=177/12\vec b\cdot\vec c = 177/12.

So cos⁡θ=(177/12)/15=177/180\cos\theta = (177/12)/15 = 177/180.

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