Ryabushko A.P. IDZ 3.1 version 10

IDZ - 3.1 No. 1.10. Given four points A1(6;8;2); A2(5;4;7); A3(2;4;7); A4(7;3;7). It is necessary to create equations:

a) equation of the plane A1A2A3; b) equation of line A1A2; c) equation of straight line A4M perpendicular to plane A1A2A3; d) equation of straight line A3N parallel to straight line A1A2; e) equation of a plane passing through point A4 and perpendicular to straight line A1A2; f) calculate the sine of the angle between straight line A1A4 and plane A1A2A3; g) calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.

Answer:

a) To compile the equation of the plane A1A2A3, it is necessary to find the vector product of the vectors $\vec{A_1A_2}$ and $\vec{A_1A_3}$:

$$\vec{A_1A_2} = \begin{pmatrix}5-6\4-8\7-2\end{pmatrix} = \begin{pmatrix}-1\-4\5\end{pmatrix}, \quad \vec{A_1A_3} = \begin{pmatrix}2-6\4-8\7-2\end{pmatrix} = \begin{pmatrix}-4\-4\5\end{pmatrix}$$

$$\vec{n} = \vec{A_1A_2} \times \vec{A_1A_3} = \begin{pmatrix}-1\-4\5\end{pmatrix} \times \begin{pmatrix}-4\-4\5\end{pmatrix} = \begin{pmatrix}-5\-9\16\end{pmatrix}$$

Thus, the equation of the plane A1A2A3 has the form:

$$-5x - 9y + 16z + d = 0$$

Substituting the coordinates of point A1, we find the value of the constant d:

$$-5\cdot6 - 9\cdot8 + 16\cdot2 + d = 0 \quad \Rightarrow \quad d = 55$$

Thus, the equation of the plane A1A2A3 is:

$$-5x - 9y + 16z + 55 = 0$$

b) The equation of straight line A1A2 can be written in parametric form:

$$x = 6 - t, \quad y = 8 - 4t, \quad z = 2 + 5t$$

d) The equation of straight line A3N can be written in parametric form:

$$x = 2 + s, \quad y = 4, \quad z = 7 - 3s$$

e) The equation of a plane passing through point A4 and perpendicular to line A1A2 has the form:

$$\vec{n} \cdot (\vec{r} - \vec{A_4}) = 0$$

where $\vec{n}$ is the direction vector of the line passing through points A1 and A2, equal to

$$\vec{n} = \begin{pmatrix}-1\-4\5\end{pmatrix}$$

Then the equation of the desired plane has the form:

$$-x - 4y + 5z + d = 0$$

Substituting the coordinates of point A4, we find the value of the constant d:

$$-7 - 12 + 35 +d = 0 \quad \Rightarrow \quad d = -16$$

Thus, the equation of a plane passing through point A4 and perpendicular to straight line A1A2:

$$-x - 4y + 5z - 16 = 0$$

f) To calculate the sine of the angle between the line A1A4 and the plane A1A2A3, it is necessary to find the projection of the vector $\vec{A_1A_4}$ onto the vector normal to the plane A1A2A3:

$$\vec{A_1A_4} = \begin{pmatrix}7-6\3-8\7-2\end{pmatrix} = \begin{pmatrix}1\-5\5\end{pmatrix}, \quad \vec{n} = \begin{pmatrix}-5\-9\16\end{pmatrix}$$

$$\sin \alpha = \frac{|\vec{A_1A_4} \cdot \vec{n}|}{|\vec{A_1A_4}| \cdot |\vec{n}|} = \frac{|(1)\cdot(-5) + (-5)\cdot(-9) + (5)\cdot(16)|}{\sqrt{1^2 + (-5)^2 + 5^2} \cdot \sqrt{(-5)^2 + (-9)^2 + 16^2}} \approx 0.919$$

g) To calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3, it is necessary to find the angle between the normal vectors of these planes:

$$\vec{n_{Oxy}} = \begin{pmatrix}0\0\1\end{pmatrix}, \quad \vec{n_{A1A2A3}} = \begin{pmatrix}-5\-9\16\end{pmatrix}$$

$$\cos \beta = \frac{\vec{n_{Oxy}} \cdot \vec{n_{A1A2A3}}}{|\vec{n_{Oxy}}| \cdot |\vec{n_{A1A2A3}}|} = \frac{(0)\cdot(-5) + (0)\cdot(-9) + (1)\cdot(16)}{\sqrt{0^2 + 0^2 + 1^2} \cdot \sqrt{(-5)^2 + (-9)^2 + 16^2}} \approx 0.784$$

No. 2.10. Write an equation of the plane in “segments” if it passes through the point M(6;-10;1) and cuts off the segment a=–3 on the Ox axis; on the Oz axis - segment c=2.

Answer:

The plane passes through the point M(6;-10;1), so its equation has the form:

$$ax + by + cz + d = 0$$

The segment on the Ox axis cut off by the plane has a length of 3, so the points of intersection of the plane with Ox are at a distance of 3 from each other. Thus, the coordinates of these points are -3 and 0. Similarly, the points of intersection of the plane with Oz are at a distance of 2 from each other, so their coordinates are 0 and 2.

Thus, the equation of the plane in “segments” has the form:

$$\frac{x}{3} - \frac{y}{10} + \frac{z}{2} - 1 = 0$$

No. 3.10. At what value of A is the plane Ax + 3y– 5z + 1 = 0 parallel to the straight line passing through the points (1;2;3) and (4;

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In order for the plane to be parallel to the line passing through the point M(x₀;y₀;z₀) and having a direction vector $\vec{v} = (a_1;b_1;c_1)$, it is necessary that the normal vector of the plane be perpendicular to the direction vector of the line . The normal vector of the plane has coordinates (A; 3; -5). Thus, the condition for a plane to be parallel to a straight line is written as:$$Aa_1 + 3b_1 - 5c_1 = 0$$Let us express A from this equation:$$A = \frac{5c_1 - 3b_1}{a_1}$$Thus, the plane will be parallel of a given straight line with a value of A equal to $\frac{5c_1 - 3b_1}{a_1}$.

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