vectors and points in skew lines given parametric equations

**jnow2** · October 12th 2011, 02:03 AM

Having trouble converting parametric equations to skew line equations!

line 1- x=1+2r, y=1+2r, z= -3+r

line 2- x=-2, y=s, z =2+s

a.) Show that these lines are perpendicular to vector v=
[1
-2
2]

**Plato** · October 12th 2011, 02:39 AM

Originally Posted by jnow2

line 1- x=1+2r, y=1+2r, z= -3+r
line 2- x=-2, y=s, z =2+s
a.) Show that these lines are perpendicular to vector v=
[1
-2
2]

Show the vectors $<2,2,1>~\&~<0,1,1>$ perpendicular to the given vector $<1,-2,2>$ .

**jnow2** · October 12th 2011, 03:17 AM

Thankyou! but cross product for the vector of line 2 keeps equaling 4 and not 0?

**Plato** · October 12th 2011, 03:42 AM

Originally Posted by jnow2

Thankyou! but cross product for the vector of line 2 keeps equaling 4 and not 0?

It is dot product, not cross product.
$<2,2,1>\cdot<1,-2,2>=2(1)+(2)(-2)+(1)(2)=0$

**HallsofIvy** · October 12th 2011, 03:42 AM

Cross product? You mean dot product, don't you? $<0, 1, 1>\cdot<1, -2, 2>= 0(1)+ 1(-2)+ 1(2)= -2+ 2= 0$
You must have dropped the "-" from -2.

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