$\begin{aligned}
& \overrightarrow{\mathrm{c}}=3 \overrightarrow{\mathrm{a}}+6 \overrightarrow{\mathrm{b}}+9(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \
& \sin ^{-1}\left(\frac{\sqrt{65}}{9}\right) \Rightarrow \sin \theta=\frac{\sqrt{65}}{9} \Rightarrow \cos \theta=\frac{4}{9} \
& \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=3|\overrightarrow{\mathrm{a}}|^2+6 \vec{a} \cdot \vec{b}=3+\frac{6 \cdot 4}{9}=\frac{51}{9} \
& \overrightarrow{\mathrm{c}} \cdot \vec{a}=3 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+6|\overrightarrow{\mathrm{~b}}|^2=\frac{3 \cdot 4}{9}+6=\frac{22}{3} \
& \therefore 9(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \vec{b})=51-22=29
\end{aligned}$
Let the angle θ, 0 lt θ lt π 2 between two unit vectors a and b be sin ^-1 ( √65 9 ). If the vector overrightarrow =3 a+6 ~b+9( a × b), then the…
JEE Main 2025 — Mathematics Vectors & 3D Geometry
Let the angle θ,0<θ<2π between two unit vectors a^ and b^ be sin−1(965). If the vector c=3a^+6 b^+9(a^×b^), then the value of 9(c⋅a^)−3(c⋅b^) is
Held on 7 Apr 2025 · Verified 6 Jul 2026.
Options
- A
31
- B
27
- C
29
- D
24
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