I have a problem where i need to find an vector $\vec{p}$, and all i have is angles with Ox and Oy, and I know that angle with Oz is obtuse. Professor in our materials used the formula ${\cos^2{\alpha}}+{\cos^2{\beta}}+{\cos^2{\gamma}}=1$ with $\alpha, \beta, \gamma$ being angles between the vector $\vec{p}$ and Ox, Oy Oz respectivly. Another thing i didn't understand why unit vector $\vec{p_0}$ was set to ($\cos{\alpha}, \cos{\beta}, \cos{\gamma}$) when vector $\vec{p}$ was being being calculated. Where was this formula derived from, and why is unit vector as is?
Finding angle between vector and Oz axis if we have angles with Ox and Oy
analytic geometrylinear algebra
Related Solutions
The volume satisfies $V=|\det D|$ where $D$ is the matrix $$\pmatrix{a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&b_2&c_3}$$ where $\mathbf{a}=(a_1,a_2,a_3)$ is the vector corresponding to $a$ etc. Thus $$V^2=\det(DD^t)=\det \pmatrix{\mathbf{a}\cdot\mathbf{a} &\mathbf{a}\cdot\mathbf{b} &\mathbf{a}\cdot\mathbf{c}\\ \mathbf{b}\cdot\mathbf{a} &\mathbf{b}\cdot\mathbf{b} &\mathbf{b}\cdot\mathbf{c}\\ \mathbf{c}\cdot\mathbf{a} &\mathbf{c}\cdot\mathbf{b} &\mathbf{c}\cdot\mathbf{c}}.$$
Now you can express $V^2$ in terms of $a$, $b$, $c$ and the various cosines.
Cosine of the angle between the diagonal vectors $\vec{v},\vec{w}$ of a parallelogram also depends on the direction of the vectors. $\vec{v}\centerdot\vec{w}$ can be negative or positive. What you teacher wants seems to be the acute angle between the diagonals. In that case, the correct formula would be
$$\cos\theta=\left|\frac{\vec{v}\centerdot\vec{w}}{||\vec{v}||||\vec{w}||}\right|.$$
How does the formula $\cos\theta=\frac{\vec{v}\centerdot\vec{w}}{||\vec{v}||||\vec{w}||}$ depend on the direction of the diagonal vectors?
From the diagram, $$\theta+\theta'=\pi\quad(\theta,\theta'>0).$$ So, $\cos\theta$ and $\cos\theta'$ must have opposite signs. In (1), $\theta$ is the angle between the diagonal vectors. We have $$\cos\theta=\frac{\vec{v}\centerdot\vec{w}}{||\vec{v}||||\vec{w}||}$$ and in (2), $\theta'$ is the angle between the diagonal vectors. We get $$\cos\theta'=\frac{\vec{v}\centerdot(\vec{-w})}{||\vec{v}||||\vec{-w}||}=-\left(\frac{\vec{v}\centerdot\vec{w}}{||\vec{v}||||\vec{w}||}\right)=-\cos\theta.$$
Therefore, $\cos\theta$ and $\cos\theta'$ have the same magnitudes but different signs. If $\theta$ is acute, $\cos\theta>0$ and $\cos\theta'<0$. Then, $$\cos\theta=\left|\frac{\vec{v}\centerdot\vec{w}}{||\vec{v}||||\vec{w}||}\right|\quad\left(0<\theta<\frac{\pi}{2}\right).$$ In case $\theta'$ is acute, we will have $\cos\theta'>0$ and $\cos\theta<0$. Then, $$\cos\theta'=\left|\frac{\vec{v}\centerdot\vec{w}}{||\vec{v}||||\vec{w}||}\right|\quad\left(0<\theta'<\frac{\pi}{2}\right).$$
Best Answer
If $u = (x, y, z)$ is a unit vector, and $i,j,k$ are the unit vector along $Ox, Oy, Oz$, then
$\cos \alpha = u \cdot i = x $
$\cos \beta = u \cdot j = y $
$\cos \gamma = u \cdot k = z $
Since $x^2 + y^2 + z^2 = 1 $ because $u$ is a unit vector, it follows that
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $