Let the angle θ , 0 < θ < 2 π between two unit vectors a ^ and b ^ be sin − 1 ( 9 65 ) . If the vector c = 3 a ^ + 6 b ^ + 9 ( a ^ × b ^ ) , then the value of 9 ( c ⋅ a ^ ) − 3 ( c ⋅ b ^ ) is

c = 3 a + 6 b + 9 ( a × b ) sin − 1 ( 9 65 ) ⇒ sin θ = 9 65 ⇒ cos θ = 9 4 c ⋅ a = 3∣ a ∣ 2 + 6 a ⋅ b = 3 + 9 6 ⋅ 4 = 9 51 c ⋅ a = 3 a ⋅ b + 6∣ b ∣ 2 = 9 3 ⋅ 4 + 6 = 3 22 ∴ 9 ( c ⋅ a ) − 3 ( c ⋅ b ) = 51 − 22 = 29

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MathematicsVectorDot Product & Its Application ( Projection etc.)Easy2 minPYQ_2025

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Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\overset{a}{^}$ and $\hat{b}$ be $sin^{- 1} (\frac{65}{9})$ . If the vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b}),$ then the value of $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$ is

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Solution:

$c = 3 a + 6 b + 9 (a \times b) sin^{- 1} (\frac{65}{9}) \Rightarrow sin θ = \frac{65}{9} \Rightarrow cos θ = \frac{4}{9} c \cdot a = 3∣ a ∣^{2} + 6 a \cdot b = 3 + \frac{6 \cdot 4}{9} = \frac{51}{9} c \cdot a = 3 a \cdot b + 6∣ b ∣^{2} = \frac{3 \cdot 4}{9} + 6 = \frac{22}{3} ∴ 9 (c \cdot a) - 3 (c \cdot b) = 51 - 22 = 29$

Stream:JEESubject:MathematicsTopic:VectorSubtopic:Dot Product & Its Application ( Projection etc.)

⏱ 2mℹ️ Source: PYQ_2025

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