IDZ Ryabushko 3.1 Option 3

  1. Given four points A1(3;5;4); A2(5;8;3); A3(1;2;–2); A4(–1;0;2). It is necessary to create equations:
    1. Planes A1A2A3;
    2. Direct A1A2;
    3. Straight line A4M, perpendicular to plane A1A2A3;
    4. Line A3N parallel to line A1A2;
    5. The plane passing through point A4, perpendicular to straight line A1A2;
    You also need to calculate:
    1. Sine of the angle between straight line A1A4 and plane A1A2A3;
    2. Cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.
  2. It is necessary to find the distance from a point to a plane.
  3. It is necessary to create an equation for a straight line passing through the point M(1;–3;3) and forming angles with the coordinate axes, respectively, of 60; 45 and 120.
    1. In order to find the equation of the plane A1A2A3, it is necessary to find the vector product of its two direction vectors:

      $$\vec{a_1} = \overrightarrow{A_1A_2} = \begin{pmatrix} 5-3 \\ 8-5 \\ 3-4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \ \vec{a_2} = \overrightarrow{A_1A_3} = \begin{pmatrix} 1-3 \\ 2-5 \\ -2-4 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \\ -6 \end{pmatrix}.$$

      Thus, $$\vec{n} = \vec{a_1} \times \vec{a_2} = \begin{pmatrix} 18 \\ 8 \\ -6 \end{pmatrix}.$$

      The plane equation can be written as:

      $$18x + 8y - 6z + d = 0.$$

      In order to find $d$, we substitute the coordinates of point $A_1$:

      $$18 \cdot 3 + 8 \cdot 5 - 6 \cdot 4 + d = 0 \Rightarrow d = -6.$$

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

      $$18x + 8y - 6z - 6 = 0.$$

      In order to find the equation of straight line A1A2, you need to find its direction vector:

      $$\vec{b} = \overrightarrow{A_1A_2} = \begin{pmatrix} 5-3 \\ 8-5 \\ 3-4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}.$$

      Thus, the equation of straight line A1A2 has the form:

      $$\begin{cases} x = 3 + 2t \\ y = 5 + 3t \\ z = 4 - t \end{p>

      In order to find the equation of the straight line A4M perpendicular to the plane A1A2A3, it is necessary to find the direction vector of this straight line. The direction vector will be directed along the vector perpendicular to the plane A1A2A3, and we have already found this vector when solving problem a). Thus, the direction vector of straight line A4M is equal to:

      $$\vec{c} = \vec{n} = \begin{pmatrix} 18 \\ 8 \\ -6 \end{pmatrix}.$$

      Considering that the point $M$ has coordinates $(1;-3;3)$, the equation of straight line A4M has the form:

      $$\begin{cases} x = 1 + 18t \\ y = -3 + 8t \\ z = 3 - 6t \end{cases}.$$

      In order to find the equation of straight line A3N parallel to straight line A1A2, it is necessary to find its direction vector. The direction vector of this line will be collinear to the vector directed along straight line A1A2, that is:

      $$\vec{d} = \overrightarrow{A_1A_2} = \begin{pmatrix} 5-3 \\ 8-5 \\ 3-4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}.$$

      Thus, the equation of straight line A3N has the form:

      $$\begin{cases} x = 1 - 2t \\ y = 2 - 3t \\ z = -2 - t \end{cases}.$$

      In order to find the equation of a plane passing through point A4 and perpendicular to line A1A2, it is necessary to find its normal vector. The normal vector of the plane will be collinear to the vector product of the vector directed along straight line A1A2 and the vector directed from point A4 to the point of intersection of straight line A1A2 with the plane A1A2A3. Thus, the normal vector of the plane is:

      $$\vec{m} = \vec{b} \times (\vec{a_1} \times \vec{b}) = \begin{pmatrix} -10 \\ 14 \\ -6 \end{pmatrix}.$$

      Considering that point A4 has coordinates $(-1;0;2)$, the plane equation has the form:

      $$-10x + 14y - 6z + d = 0.$$

      In order to find $d$, we substitute the coordinates of point A4:

      $$-10 \cdot (-1) + 14 \cdot 0 - 6 \cdot 2 + d = 0 \Rightarrow d = -2.$$

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

      $$-10x + 14y - 6z - 2 = 0.$$

      To calculate the sine of the angle between straight line A1A4 and plane A1A2A3

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