First, we write b → \overrightarrow{b} b as a linear combination of a → \overrightarrow{a} a and j → \overrightarrow{j} j since b → \overrightarrow{b} b is a rotation of a → \overrightarrow{a} a about the y-axis.
b ⃗ = λ a ⃗ + μ j ^ = λ ( − i ^ + 2 j ^ + k ^ ) + μ j ^ = − λ i ^ + ( 2 λ + μ j ^ ) + λ k ^ \vec{b}=\lambda \vec{a}+\mu \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})+\mu \hat{j}=-\lambda \hat{i}+(2 \lambda+\mu \hat{j})+\lambda \hat{k} b = λ a + μ j ^ = λ ( − i ^ + 2 j ^ + k ^ ) + μ j ^ = − λ i ^ + ( 2 λ + μ j ^ ) + λ k ^
b → \overrightarrow{b} b is orthogonal to a → \overrightarrow{a} a due to the right angle rotation, so b → ⋅ a → = 0 \overrightarrow{b} \cdot \overrightarrow{a} = 0 b ⋅ a = 0 .
This implies :
a m p ; ( − i ^ + 2 j ^ + k ^ ) ⋅ ( − λ i ^ + ( 2 λ + μ ) j ^ + λ k ^ ) = 0 a m p ; λ + 2 ( 2 λ + μ ) + λ = 0 ⇒ 6 λ + 2 μ = 0 ⇒ μ + 3 λ = 0 \begin{aligned} & (-\hat{i}+2 \hat{j}+\hat{k}) \cdot(-\lambda \hat{i}+(2 \lambda+\mu) \hat{j}+\lambda \hat{k})=0 \\\\ & \lambda+2(2 \lambda+\mu)+\lambda=0 \Rightarrow 6 \lambda+2 \mu=0 \Rightarrow \mu+3 \lambda=0\end{aligned} am p ; ( − i ^ + 2 j ^ + k ^ ) ⋅ ( − λ i ^ + ( 2 λ + μ ) j ^ + λ k ^ ) = 0 am p ; λ + 2 ( 2 λ + μ ) + λ = 0 ⇒ 6 λ + 2 μ = 0 ⇒ μ + 3 λ = 0
∴ \therefore ∴ b ⃗ = λ a ⃗ − 3 λ j ^ = λ ( − i ^ + 2 j ^ + k ^ ) − 3 λ j ^ = λ ( − i ^ − j ^ + k ^ ) \vec{b}=\lambda \vec{a}-3 \lambda \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})-3 \lambda \hat{j}=\lambda(-\hat{i}-\hat{j}+\hat{k}) b = λ a − 3 λ j ^ = λ ( − i ^ + 2 j ^ + k ^ ) − 3 λ j ^ = λ ( − i ^ − j ^ + k ^ )
The magnitude of b → \overrightarrow{b} b is the same as the magnitude of a → \overrightarrow{a} a because a rotation doesn't change the magnitude of a vector. This gives us :
∣ b ⃗ ∣ = 3 ∣ λ ∣ = 6 [ ∵ ∣ a ∣ = 6 ] ⇒ ∣ λ ∣ = 2 ⇒ λ ≠ 2 |\vec{b}|=\sqrt{3}|\lambda|=\sqrt{6}[\because|a|=\sqrt{6}] \Rightarrow|\lambda|=\sqrt{2} \Rightarrow \lambda \neq \sqrt{2} ∣ b ∣ = 3 ∣ λ ∣ = 6 [ ∵ ∣ a ∣ = 6 ] ⇒ ∣ λ ∣ = 2 ⇒ λ = 2 λ \lambda λ angle between b b b and y y y -axis is not acute. λ = − 2 \lambda=-\sqrt{2} λ = − 2
Thus, we have :
b → = − 2 ( − i ^ − j ^ + k ^ ) = 2 i ^ + 2 j ^ − 2 k ^ \overrightarrow{b} = -\sqrt{2}(-\hat{i} - \hat{j} + \hat{k}) = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k} b = − 2 ( − i ^ − j ^ + k ^ ) = 2 i ^ + 2 j ^ − 2 k ^
Then, we find the vector 3 a → + 2 b → 3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} 3 a + 2 b :
3 a → + 2 b → = 3 ( − i ^ + 2 j ^ + k ^ ) + 2 ( 2 i ^ + 2 j ^ − 2 k ^ ) 3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} = 3(-\hat{i} + 2\hat{j} + \hat{k}) + \sqrt{2}(\sqrt{2}\hat{i} + \sqrt{2}\hat{j} - \sqrt{2}\hat{k}) 3 a + 2 b = 3 ( − i ^ + 2 j ^ + k ^ ) + 2 ( 2 i ^ + 2 j ^ − 2 k ^ ) = − 3 i ^ + 6 j ^ + 3 k ^ + 2 i ^ + 2 j ^ − 2 k ^ = − i ^ + 8 j ^ + k ^ = -3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} - 2\hat{k} = -\hat{i} + 8\hat{j} + \hat{k} = − 3 i ^ + 6 j ^ + 3 k ^ + 2 i ^ + 2 j ^ − 2 k ^ = − i ^ + 8 j ^ + k ^
The projection of this vector onto c → \overrightarrow{c} c is given by the dot product divided by the magnitude of c → \overrightarrow{c} c :
( − i ^ + 8 j ^ + k ^ ) ⋅ ( 5 i ^ + 4 j ^ + 3 k ^ ) 5 2 + 4 2 + 3 2 = − 5 + 32 + 3 50 = 30 5 2 = 3 2 \frac{(-\hat{i} + 8\hat{j} + \hat{k}) \cdot (5\hat{i} + 4\hat{j} + 3\hat{k})}{\sqrt{5^2 + 4^2 + 3^2}} = \frac{-5 + 32 + 3}{\sqrt{50}} = \frac{30}{5\sqrt{2}} = 3\sqrt{2} 5 2 + 4 2 + 3 2 ( − i ^ + 8 j ^ + k ^ ) ⋅ ( 5 i ^ + 4 j ^ + 3 k ^ ) = 50 − 5 + 32 + 3 = 5 2 30 = 3 2
So the projection of 3 a → + 2 b → 3\overrightarrow{a} + \sqrt{2}\overrightarrow{b} 3 a + 2 b onto c → \overrightarrow{c} c is 3 2 3\sqrt{2} 3 2 which is option D.