Question

Find the direction cosines of the line joining the points Plt,
Q(-2, 1,-8).
OR
Find the value of p for which the following lines are perpendicular:
1-X 2y – 14 2-3, 1-x _ y - 5 - 6 - 2
3
20 2. 3p 1
5
​

Answer 1

Answer:

Note that the converse holds as well. If  u⇀=kv⇀  for some scalar  k , then either  u⇀  and  v⇀  have the same direction  (k>0)  or opposite directions  (k<0) , so  u⇀  and  v⇀  are parallel. Therefore, two nonzero vectors  u⇀  and  v⇀  are parallel if and only if  u⇀=kv⇀  for some scalar  k . By convention, the zero vector  0⇀  is considered to be parallel to all vectors.

Figure  11.5.1 : Vector  v⇀  is the direction vector for  PQ−−⇀ .

As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallel vector, which we call the direction vector (Figure  11.5.1 ). Let  L  be a line in space passing through point  P(x0,y0,z0) . Let  v⇀=⟨a,b,c⟩  be a vector parallel to  L . Then, for any point on line  Q(x,y,z) , we know that  PQ−−⇀  is parallel to  v⇀ . Thus, as we just discussed, there is a scalar,  t , such that  PQ−−⇀=tv⇀ , which gives

PQ−−⇀⟨x−x0,y−y0,z−z0⟩⟨x−x0,y−y0,z−z0⟩=tv⇀=t⟨a,b,c⟩=⟨ta,tb,tc⟩.(11.5.3)

Using vector operations, we can rewrite Equation  11.5.3  

⟨x−x0,y−y0,z−z0⟩⟨x,y,z⟩−⟨x0,y0,z0⟩⟨x,y,z⟩r⇀=⟨ta,tb,tc⟩=t⟨a,b,c⟩=⟨x0,y0,z0⟩r⇀o+t⟨a,b,c⟩v⇀.

 

Setting  r⇀=⟨x,y,z⟩  and  r⇀0=⟨x0,y0,z0⟩ , we now have the vector equation of a line:

r⇀=r⇀0+tv⇀.(11.5.4)

Equating components, Equation  11.5.4  shows that the following equations are simultaneously true:  x−x0=ta,y−y0=tb,  and  z−z0=tc.  If we solve each of these equations for the component variables  x,y,  and  z , we get a set of equations in which each variable is defined in terms of the parameter  t  and that, together, describe the line. This set of three equations forms a set of parametric equations of a line:

x=x0+ta

 

y=y0+tb

 

z=z0+tc.

 

If we solve each of the equations for  t  assuming  a,b , and  c  are nonzero, we get a different description of the same line:

x−x0ay−y0bz−z0c=t=t=t.

 

Because each expression equals  t , they all have the same value. We can set them equal to each other to create symmetric equations of a line:

x−x0a=y−y0b=z−z0c.

Step-by-step explanation: