Let a = - i + 2 j + 2 k$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ , b = 8 i + 7 j - 3 k$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$ and c$\vec{c}$ be a vector such that a × c = b$\vec{a} \times \vec{c} = \vec{b}$ . If c. ( i + j + k) = 4$\vec{c}. (\hat{i} + \hat{j} + \hat{k}) = 4$ , then | a + c|²$|\vec{a} + \vec{c}|^2$ is equal to:
Let the unknown vector be c = c₁ i + c₂ j + c₃ k$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.
Given a = - i + 2 j + 2 k$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and b = 8 i + 7 j - 3 k$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$.
Compute the cross product a × c$\vec{a} \times \vec{c}$:
Whenever a × c = b$\vec{a} \times \vec{c} = \vec{b}$ and a dot product constraint is given, simply expand c$\vec{c}$ algebraically as (c₁, c₂, c₃)$(c_1, c_2, c_3)$. The cross product and dot product create an easily solvable 3x3 linear system.
Let a$\vec{a}$ and b$\vec{b}$ be two vectors such that | a| = 1, | b| = 4$|\vec{a}| = 1, |\vec{b}| = 4$ and a · b = 2$\vec{a} \cdot \vec{b} = 2$. If c = (2 a × b) - 3 b$\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ and the angle between b$\vec{b}$ and c$\vec{c}$ is α$\alpha$, then 192 ²α$192\sin^2\alpha$ is equal to
Numerical Answer.Answer: 48 to 48
Solution
Core Logic
b · c = b · ((2 a × b) - 3 b)$$\vec{b} \cdot \vec{c} = \vec{b} \cdot ((2\vec{a} \times \vec{b}) - 3\vec{b})$$|b||c| α = 2( b · ( a × b)) - 3|b|²$$|b||c|\cos\alpha = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3|b|^2$$
Since b · ( a × b) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$, we have |b||c| α = -3|b|²$|b||c|\cos\alpha = -3|b|^2$.
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