Question

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

Answer 1

Answer on Question #46072 – Math – Analytic Geometry

Question. Find the angle between the lines

$P:\ x=1,\qquad z-y=0$

and

$Q:\ 2x-y=-1,\qquad z=1.$

Solution. Let us find vectors $p$ and $q$ which are parallel to the lines $P$ and $Q$.

By assumption the line $p$ is the intersection of two planes $x=1$ and $z-y=0$ having the following normal vectors:

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

Hence $p$ must be orthogonal to both $a_{1}$ and $a_{2}$. Therefore we can take $p$ to be the cross-product $a_{1}\times a_{2}$ of these vectors:

$p$ $=a_{1}\times a_{2}=(1,0,0)\times(0,-1,1)$

$=\left(\left|\begin{array}[]{cc}0&0\\

-1&1\end{array}\right|,\left|\begin{array}[]{cc}0&1\\

1&0\end{array}\right|\,\left|\begin{array}[]{cc}1&0\\

0&-1\end{array}\right|\right)$

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

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

Analogously, the line $q$ is the intersection of two planes $2x-y=-1$ and $z=1$ having normal vectors:

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

and we can assume that

$q$ $=b_{1}\times b_{2}=(2,-1,0)\times(0,0,1)$

$=\left(\left|\begin{array}[]{cc}-1&0\\

0&1\end{array}\right|\,\left|\begin{array}[]{cc}0&2\\

1&0\end{array}\right|\,\left|\begin{array}[]{cc}2&-1\\

0&0\end{array}\right|\right)$

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

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

To find the angle $\phi$ between the vectors $p$ and $q$ notice that the scalar product $(p,q)$ of these vectors can be computed in two distinct ways. On the one hand, it is the sum of products of the corresponding coordinates:

$(p,q)=0\cdot(-1)+(-1)\cdot(-2)+(-1)\cdot 0=2.$

On the other hand,

$(p,q)=|p|\cdot|q|\cdot\cos\phi,$

whence

$\cos\phi=\frac{(p,q)}{|p|\cdot|q|}.$

The lengths of these vectors are

$|p|=\sqrt{0^{2}+(-1)^{2}+(-1)^{2}}=\sqrt{2},\qquad|q|=\sqrt{(-1)^{2}+(-2)^{2}+0^{2}}=\sqrt{5},$

therefore

$\cos\phi=\frac{(p,q)}{|p|\cdot|q|}=\frac{2}{\sqrt{2}\cdot\sqrt{5}}=\frac{\sqrt{2}}{\sqrt{5}}=\sqrt{\frac{2}{5}}=\sqrt{0.4}\approx 0.6325.$

Hence

$\phi=\arccos\sqrt{0.4}\approx\arccos(0.6325)\approx 50.77{}^{\circ}.$

Answer. The angle between the lines is

$\arccos\sqrt{0.4}\approx 50.77{}^{\circ}.$

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