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Question: Let $\vec{\alpha} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{\beta} = \hat{i} + 2\hat{j} - 4\hat{k}...

Let α⃗=4i^+3j^+5k^\vec{\alpha} = 4\hat{i} + 3\hat{j} + 5\hat{k} and β⃗=i^+2j^−4k^\vec{\beta} = \hat{i} + 2\hat{j} - 4\hat{k}. Let β1⃗\vec{\beta_1} be parallel to α⃗\vec{\alpha} and β2⃗\vec{\beta_2} be perpendicular to α⃗\vec{\alpha}. If β⃗=β1⃗+β2⃗\vec{\beta} = \vec{\beta_1} + \vec{\beta_2}, then the value of 5β2⃗⋅(i^+j^+k^)5\vec{\beta_2} \cdot (\hat{i} + \hat{j} + \hat{k}) is

A

9

B

7

C

6

D

11

Answer

7

Explanation

Solution

Given β⃗=β1⃗+β2⃗\vec{\beta} = \vec{\beta_1} + \vec{\beta_2} where β1⃗∥α⃗\vec{\beta_1} \parallel \vec{\alpha} and β2⃗⊥α⃗\vec{\beta_2} \perp \vec{\alpha}. β1⃗\vec{\beta_1} is the projection of β⃗\vec{\beta} onto α⃗\vec{\alpha}, β1⃗=β⃗⋅α⃗∣α⃗∣2α⃗\vec{\beta_1} = \frac{\vec{\beta} \cdot \vec{\alpha}}{|\vec{\alpha}|^2} \vec{\alpha}. β2⃗=β⃗−β1⃗\vec{\beta_2} = \vec{\beta} - \vec{\beta_1}. We need 5β2⃗⋅(i^+j^+k^)5\vec{\beta_2} \cdot (\hat{i} + \hat{j} + \hat{k}). This is 5(β⃗−β⃗⋅α⃗∣α⃗∣2α⃗)⋅(i^+j^+k^)5\left(\vec{\beta} - \frac{\vec{\beta} \cdot \vec{\alpha}}{|\vec{\alpha}|^2} \vec{\alpha}\right) \cdot (\hat{i} + \hat{j} + \hat{k}). This expands to 5(β⃗⋅(i^+j^+k^)−β⃗⋅α⃗∣α⃗∣2(α⃗⋅(i^+j^+k^)))5\left(\vec{\beta} \cdot (\hat{i} + \hat{j} + \hat{k}) - \frac{\vec{\beta} \cdot \vec{\alpha}}{|\vec{\alpha}|^2} (\vec{\alpha} \cdot (\hat{i} + \hat{j} + \hat{k}))\right). Calculating the dot products: β⃗⋅α⃗=−10\vec{\beta} \cdot \vec{\alpha} = -10, ∣α⃗∣2=50|\vec{\alpha}|^2 = 50, β⃗⋅(i^+j^+k^)=−1\vec{\beta} \cdot (\hat{i} + \hat{j} + \hat{k}) = -1, α⃗⋅(i^+j^+k^)=12\vec{\alpha} \cdot (\hat{i} + \hat{j} + \hat{k}) = 12. Substituting these values: 5(−1−−1050⋅12)=5(−1+125)=5(75)=75\left(-1 - \frac{-10}{50} \cdot 12\right) = 5\left(-1 + \frac{12}{5}\right) = 5\left(\frac{7}{5}\right) = 7.