Final answer:
The line perpendicular to both the vector u=<1,1,-2> and the z-axis through the point (-2,5,3) is found by taking the cross product to determine the direction vector. Its parametric equations are x=-2-t, y=5+2t, z=3-t.
Step-by-step explanation:
To give a parametric representation of the line through the point (-2,5,3) that is perpendicular to both u=<1,1,-2> and the z-axis, we need to find a direction vector for this line that is perpendicular to both u and the z-axis. Since u and the z-axis are not parallel, we can find such a vector by taking the cross product of u and a vector representing the z-axis, which can be taken as k=<0,0,1>.
First, we compute the cross product of u and k:
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- u x k = | i j k |
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- | 1 1 -2 |
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- | 0 0 1 |
This results in v=<1*(-1)-(-2)*0, -1*(1)-(-2)*0, 1*0-1*1> which simplifies to v=<-1, 2, -1>.
The parametric equations for the line can now be written using the vector v and the given point (-2,5,3) as:
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- x = -2 - t
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- y = 5 + 2t
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- z = 3 - t
Where t is a parameter that can take any real number value.