The problem asks us to find the projection of the straight line (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)\frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1} onto the plane x+y+2z=6x+y+2z=6.
Work/Calculations
Step 1: Find the Direction Vector of the Line
The direction vector of the line is given by [2,3,-1][2, 3, -1].
Step 2: Find the Normal Vector of the Plane
The normal vector of the plane x+y+2z=6x+y+2z=6 is [1,1,2][1, 1, 2].
Step 3: Find the Projection Vector
The projection of the direction vector of the line onto the plane is given by “Projection”=”Direction Vector”-((“Direction Vector”*”Normal Vector”)/(“Normal Vector”*”Normal Vector”))xx”Normal Vector”\text{Projection} = \text{Direction Vector} – \left( \frac{\text{Direction Vector} \cdot \text{Normal Vector}}{\text{Normal Vector} \cdot \text{Normal Vector}} \right) \times \text{Normal Vector}
The projected line will pass through a point on the original line and will have the direction vector as the projection vector. Taking the point (1,1,-1)(1, 1, -1) on the original line, the equation of the projected line is:
The projection of the given straight line (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)\frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1} onto the plane x+y+2z=6x+y+2z=6 is the line (x-1)/(1.5)=(y-1)/(2.5)=(z+1)/(-2).\frac{x-1}{1.5} = \frac{y-1}{2.5} = \frac{z+1}{-2}.