Question

Find the point of intersection of the line x/4=y=z-1 and the plane 2x+y+z=5.
Also find the angle between them.

Answer 1

Answer on Question #45075 – Math – Analytic Geometry

Find the point of intersection of the line $x/4 = y = z - 1$ and the plane $2x + y + z = 5$. Also find the angle between them.

Solution.

To find the point of intersection of the line

$$\frac{x}{4} = y = z - 1$$

and the plane

$$2x + y + z = 5,$$

we solve the system of linear equations

$$\left\{ \begin{array}{l} \frac{x}{4} = y = z - 1 \\ 2x + y + z = 5 \end{array} \right.$$

Express variables $x$ and $z$ as functions of $y$ on the basis of (1), we come to

$$x = 4y, z = y + 1.$$

Plug these expressions into the second equation (2) of the system (3):

$$2x + y + z = 5, \quad 2 * 4y + y + y + 1 = 5,$$

Collect similar terms: $10y = 4$, divide both sides by 10 and obtain $y = \frac{4}{10}$, $y = \frac{2}{5}$.

Other coordinates of this point are $x = 4y = 4 * \frac{4}{10} = \frac{8}{5}$, $z = y + 1 = \frac{2}{5} + 1 = \frac{7}{5}$.

Thus, the point of intersection of the line and plane is $(x,y,z) = \left(\frac{8}{5},\frac{2}{5},\frac{7}{5}\right)$.

Vector $N = (A, B, C) = (2, 1, 1)$ is the normal to the plane (2), vector

$\pmb{u} = (l, m, n) = (4, 1, 1)$ is the direction vector of line (1).

The angle between the line and the plane is defined as the complementary angle $\theta$ of the angle $\varphi$ between the direction vector $\pmb{u}$ of the line and the normal $\pmb{N}$ to the plane. For this angle, one has the formula $\sin(\theta) = \cos(\varphi) = \frac{Al + Bm + Cn}{\sqrt{A^2 + B^2 + C^2} \sqrt{l^2 + m^2 + n^2}} = \frac{2*4 + 1*1 + 1*1}{\sqrt{2^2 + 1^2 + 1^2} \sqrt{4^2 + 1^2 + 1^2}} = \frac{10}{\sqrt{6*18}} = \frac{10}{6\sqrt{3}} = \frac{5}{3\sqrt{3}}$, hence $\theta = 1.295$ radians or $\theta = 74.21{}^\circ$.

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