Question

Find the angle between the lines x=1 z-y=0 and 2x-y=-1,z=1

Answer 1

Answer on Question #46166 – Math – Analytic Geometry

Question. Find the angle $\alpha$ between the lines

$L_{1}:\ x=1,\qquad z-y=0$

and

$L_{2}:\ 2x-y=-1,\qquad z=1.$

Solution. First we should find the vectors $l_{1}$ and $l_{2}$ parallel to these lines. The line $L_{1}$ is the intersection of two planes $x=1$ and $z-y=0$. The normal vectors to these planes are respectively:

$n_{1}=(1,0,0),\qquad n_{2}=(0,-1,1).$

Therefore $l_{1}$ is orthogonal to both $n_{1}$ and $n_{2}$, and so it is parallel to the cross-product of $n_{1}$ and $n_{2}$. Thus we can assume that

$l_{1}$ =n_{1}\times n_{2}=(1,0,0)\times(0,-1,1)=\left(\left|\begin{array}[]{cccc}0&0\cr-1&1\end{array}\right|\cdot\left|\begin{array}[]{cccc}0&1\cr 1&0\end{array}\right|\cdot\left|\begin{array}[]{cccc}1&0\cr 0&-1\end{array}\right|\right)

$=(0\cdot 1-0\cdot(-1),\ 0\cdot 0-1\cdot 1,\ 1\cdot(-1)-0\cdot 0)$

$=(0,-1-1)\,.$

Similarly, the line $L_{2}$ is the intersection of two planes $2x-y=-1$ and $z=1$. The normal vectors to these planes are respectively:

$p_{1}=(2,-1,0),\qquad p_{2}=(0,0,1).$

Therefore $l_{2}$ can be taken to be the cross-product of $p_{1}$ and $p_{2}$:

$l_{2}$ =p_{1}\times p_{2}=(2,-1,0)\times(0,0,1)=\left(\left|\begin{array}[]{cccc}-1&0\cr 0&1\end{array}\right|\cdot\left|\begin{array}[]{cccc}0&2\cr 1&0\end{array}\right|\cdot\left|\begin{array}[]{cccc}2&-1\cr 0&0\end{array}\right|\right)

$=(-1\cdot 1-0\cdot 0,\ 0\cdot 0-1\cdot 2,\ 2\cdot 0-0\cdot(-1))$

$=(-1,-2,0).$

The cosine of the angle $\alpha$ between $l_{1}$ and $l_{2}$ can be computed by the following formula:

$\cos\alpha=\frac{(l_{1},l_{2})}{|l_{1}|\cdot|l_{2}|}.$

We have that the scalar product $(l_{1},l_{2})$ is equal to the sum of products of the corresponding coordinates:

$(l_{1},l_{2})=0\cdot(-1)+(-1)\cdot(-2)+(-1)\cdot 0=2,$

and the lengths of these vectors are

$|l_{1}|$ $=\sqrt{0^{2}+(-1)^{2}+(-1)^{2}}=\sqrt{2},$

$|l_{2}|$ $=\sqrt{(-1)^{2}+(-2)^{2}+0^{2}}=\sqrt{5}.$

Hence

$\cos\alpha=\frac{(l_{1},l_{2})}{|l_{1}|\cdot|l_{2}|}=\frac{2}{\sqrt{2}\cdot\sqrt{5}}=\frac{\sqrt{2}}{\sqrt{5}}=\sqrt{2/5}=\sqrt{0.4}.$

Therefore

$\alpha=\arccos\sqrt{0.4}\approx\arccos(0.63246)\approx 50.768{}^{\circ}.$

Answer. $\alpha=\arccos\sqrt{0.4}\approx 50.768{}^{\circ}$.