Solutions to Problem Set 5

Math 311-01/02

2-1-2018

[Lines and planes problems]

1. Find the parametric and the symmetric equations for the line which passes through the point $(3, -4,
   7)$ and is parallel to the vector $(10, 13, -2)$ .

The parametric equations are

$$x - 3 = 10 t, \quad y + 4 = 13 t, \quad z - 7 = -2 t.$$

Solve each of these equations for t and equate the results to get the symmetric equations:

$$\dfrac{x - 3}{10} = \dfrac{y + 4}{13} = \dfrac{z - 7}{-2}.\quad\halmos$$


2. Find parametric equations for the line which passes through the point $(3, -4, 6)$ and is parallel to the line

$$x - 5 = 12 t, \quad y + 8 = 3 t, \quad z = 14 t.$$

The given line is parallel to the vector $(12, 3, 14)$ , so the same is true for the line I'm constructing. The line is

$$x - 3 = 12 t, \quad y + 4 = 3 t, \quad z - 6 = 14 t.\quad\halmos$$


3. Find the parametric equation of the line which passes through the points $P(3, 2, 5)$ and $Q(-4, 0, 9)$ .

The line is parallel to $\bvec{P
   Q} = (-7, -2, 4)$ . The equation of the line can be written using either P or Q:

$$x - 3 = -7 t, \quad y - 2 = -2 t, \quad z - 5 = 4 t,$$

$$x + 4 = -7 t, \quad y = -2 t, \quad z - 9 = 4 t.\quad\halmos$$


4. Find the parametric equation of the line which passes through the point $(3, 7, -10)$ and is perpendicular to both of the vectors $(1, 1, 5)$ and $(3, -1, 2)$ .

$$(1, 1, 5) \times (3, -1, 2) = \left|\matrix{ \ihat & \jhat & \khat \cr 1 & 1 & 5 \cr 3 & -1 & 2 \cr}\right| = (7, 13, -4).$$

The line is parallel to $(7, 13,
   -4)$ . Its equation is

$$x - 3 = 7 t, \quad y - 7 = 13 t, \quad z + 10 = -4 t.\quad\halmos$$


5. Find the equation of the plane which contains the point $(10, 1, 6)$ and is perpendicular to the vector $(4, -3, 7)$ .

$$4 (x - 10) - 3 (y - 1) + 7 (z - 6) = 0, \quad\hbox{or}\quad 4 x - 3 y + 7 z = 79.\quad\halmos$$


6. Find the equation of the plane containing the points $P(1, 1, 3)$ , $Q(2, 3, -5)$ , and $R(0, 6, 1)$ .

The vectors $\bvec{P Q} = (1, 2,
   -8)$ and $\bvec{P R} = (-1, 5, -2)$ lie in the plane, so their cross product is perpendicular to the plane:

$$(1, 2, -8) \times (-1, 5, -2) = \left|\matrix{ \ihat & \jhat & \khat \cr 1 & 2 & -8 \cr -1 & 5 & -2 \cr}\right| = (36, 10, 7).$$

$$\hbox{\epsfysize=1.5in \epsffile{lines-and-planes-problems-1.eps}}$$

Using P as the point in the plane, the plane is

$$36 (x - 1) + 10 (y - 1) + 7 (z - 3) = 0 \quad\hbox{or}\quad 36 x + 10 y + 7 z = 67.\quad\halmos$$


7. Sketch the following plane by finding its x, y, and z-intercepts:

$$2 x + 3 y + z = 6.$$

Setting $y = z = 0$ gives $2 x = 6$ , or $x = 3$ .

Setting $x = z = 0$ gives $3 y = 6$ , or $y = 2$ .

Setting $x = y = 0$ gives $z = 6$ .

$$\hbox{\epsfysize=1.5in \epsffile{lines-and-planes-problems-2.eps}}\quad\halmos$$


8. Find the equation of the plane which contains the point $(5, -6, 1)$ and is parallel to the plane

$$3 x + 10 y - 7 z = 15.$$

The given plane is perpendicular to the vector $(3, 10, -7)$ . Since the planes are parallel, the plane I want is also perpendicular to the vector $(3, 10, -7)$ . the plane is

$$3 (x - 5) + 10 (y + 6) - 7 (z - 1) = 0, \quad\hbox{or}\quad 3 x + 10 y - 7 z = -52.\quad\halmos$$


The struggle to understand is our only advantage over this madness. - Ta-Nehisi Coates


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