Return to Question

Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c} \text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\ \therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\ \quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\ \Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2} \end{array}$$

a = {x1, y1, z1};
b = {x2, y2, z2};
Solve[{7 a . a + 16 a . b - 15 b . b == 0, 
   7 a . a - 30 a . b + 8 b . b == 0, 
   cosa == a . b/(Norm[a] Norm[b])}, 
  cosa, {x1, y1, z1, x2, y2, z2}] // FullSimplify

How to use MMA to directly solve the above vector equation and find the angle between vector $\mathbf{a}$ and vector $\mathbf{b}$? I wonder if there is any other easier way.

Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c} \text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\ \therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\ \quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\ \Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2} \end{array}$$

a = {x1, y1, z1};
b = {x2, y2, z2};
Solve[{7 a . a + 16 a . b - 15 b . b == 0, 
   7 a . a - 30 a . b + 8 b . b == 0, 
   cosa == a . b/(Norm[a] Norm[b])}, 
  cosa, {x1, y1, z1, x2, y2, z2}] // FullSimplify

How to use Mathematica to directly solve the above vector equation and find the angle between vector $\mathbf{a}$ and vector $\mathbf{b}$? I wonder if there is any other easier way.

Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c} \text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\ \therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\ \quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\ \Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2} \end{array}$$

Solve[{7 a^2 + 16 a b - 15 b^2 == 0, 7 a^2 - 30 a b + 8 b^2 == 0},{a.b}]

How to use MMA to directly solve the above vector equation and find the angle between vector $\mathbf{a}$ and vector $\mathbf{b}$?

Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c} \text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\ \therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\ \quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\ \Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2} \end{array}$$

a = {x1, y1, z1};
b = {x2, y2, z2};
Solve[{7 a . a + 16 a . b - 15 b . b == 0, 
   7 a . a - 30 a . b + 8 b . b == 0, 
   cosa == a . b/(Norm[a] Norm[b])}, 
  cosa, {x1, y1, z1, x2, y2, z2}] // FullSimplify

How to use MMA to directly solve the above vector equation and find the angle between vector $\mathbf{a}$ and vector $\mathbf{b}$? I wonder if there is any other easier way.

Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c} \text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\ \therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\ \quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\ \Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2} \end{array}$$

Given the vector relation $(\mathbf{a}+3 \mathbf{b}) \perp(7 \mathbf{a}-5 \mathbf{b})$, $(\mathbf{a}-4 \mathbf{b}) \perp(7 \mathbf{a}-2 \mathbf{b})$, find the angle between $\mathbf{a}$ and $\mathbf{b}$.

$$\begin{array}{c} \text{Solving process: } \qquad \because(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5\mathbf{b}) = 0,(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b}) = 0 \\ \therefore \quad 7 \mathbf{a}^{2}+16 \mathbf{a} \cdot \mathbf{b}-15 \mathbf{b}^{2} = 0,7 \mathbf{a}^{2}-30 \mathbf{a} \cdot \mathbf{b}+8 \mathbf{b}^{2} = 0\\ \quad \mathbf{a}^{2} = \mathbf{b}^{2} = 2 \mathbf{a} \cdot \mathbf{b} \\ \Rightarrow \cos \langle \mathbf{a}, \mathbf{b}\rangle = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}|\cdot \mathbf{b}|} = \frac{1}{2} \end{array}$$

Solve[{7 a^2 + 16 a b - 15 b^2 == 0, 7 a^2 - 30 a b + 8 b^2 == 0},{a.b}]

How to use MMA to directly solve the above vector equation and find the angle between vector $\mathbf{a}$ and vector $\mathbf{b}$?