Angle chase #1

Geometry Level pending

Given four non-zero vectors $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$ . The vectors $\vec{a},\vec{b}$ and $\vec{c}$ are coplanar but non- collinear pair by pair and vector $\vec{d}$ is not coplanar with vectors $\vec{a},\vec{b}$ and $\vec{c}$ and $(\hat{a}\cdot \hat{b})=(\hat{b} \cdot \hat{c})= \dfrac{1}{2} , (\hat{d}\cdot \hat{a}) = \cos \alpha$ and $(\hat{d} \cdot \hat{b}) =\cos \beta$ .

If $(\hat{d}\cdot\hat{c})= (m\cos \beta +n\cos \alpha)$ , then find $m-n$

The answer is 2.

1 solution

A Former Brilliant Member
Dec 30, 2017

Let $\hat{d}=\lambda_{1} \hat{a} +\lambda_{2} \hat{b} + \lambda_{3} (\hat{a} \times \hat{b})$

$\hat{d} \cdot \hat{a}= \lambda_{1} +\dfrac{\lambda_{2}}{2} \implies \lambda_{1} +\dfrac{\lambda_{2}}{2} = \cos \alpha$

$\hat{d} \cdot \hat{b}= \lambda_{2} +\dfrac{\lambda_{1}}{2} \implies \lambda_{2} +\dfrac{\lambda_{1}}{2} = \cos \beta$

Solving the above equations, we get:

$\lambda_{1}=\dfrac{2}{3}(2\cos \alpha - \cos \beta)$

$\lambda_{2}=\dfrac{2}{3}(2\cos \beta - \cos \alpha)$

Now, calculate $\hat{d}\cdot\hat{c}$ :

$\hat{d}\cdot\hat{c} = \dfrac{-\lambda_{1}}{2}+\dfrac{\lambda_2}{2}$

$\hat{d}\cdot\hat{c}= \cos \beta - \cos \alpha$

$\therefore \boxed{m=1}$ and $\boxed{n=-1}$

$\implies \boxed{m-n=2}$

Angle chase #1

The answer is 2.

1 solution

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