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Question:-1(e)

b^2=c^2+a^2-2ac\:Cos\left(B\right)

Find the projection of the straight line \frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1} on the plane x+y+2 z=6.

Expert Answer
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Introduction

The problem asks us to find the projection of the straight line x 1 2 = y 1 3 = z + 1 1 x 1 2 = y 1 3 = z + 1 1 (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)\frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1}x12=y13=z+11 onto the plane x + y + 2 z = 6 x + y + 2 z = 6 x+y+2z=6x+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 ] [2,3,-1][2, 3, -1][2,3,1].

Step 2: Find the Normal Vector of the Plane

The normal vector of the plane x + y + 2 z = 6 x + y + 2 z = 6 x+y+2z=6x+y+2z=6x+y+2z=6 is [ 1 , 1 , 2 ] [ 1 , 1 , 2 ] [1,1,2][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 ) × Normal Vector Projection = Direction Vector Direction Vector Normal Vector Normal Vector Normal Vector × Normal Vector “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}Projection=Direction Vector(Direction VectorNormal VectorNormal VectorNormal Vector)×Normal Vector
After substituting the values, we get:
Projection = [ 2 , 3 , 1 ] ( [ 2 , 3 , 1 ] [ 1 , 1 , 2 ] [ 1 , 1 , 2 ] [ 1 , 1 , 2 ] ) × [ 1 , 1 , 2 ] Projection = [ 2 , 3 , 1 ] [ 2 , 3 , 1 ] [ 1 , 1 , 2 ] [ 1 , 1 , 2 ] [ 1 , 1 , 2 ] × [ 1 , 1 , 2 ] “Projection”=[2,3,-1]-(([2,3,-1]*[1,1,2])/([1,1,2]*[1,1,2]))xx[1,1,2]\text{Projection} = [2, 3, -1] – \left( \frac{[2, 3, -1] \cdot [1, 1, 2]}{[1, 1, 2] \cdot [1, 1, 2]} \right) \times [1, 1, 2]Projection=[2,3,1]([2,3,1][1,1,2][1,1,2][1,1,2])×[1,1,2]
After Calculating, we get:
Projection = [ 2 , 3 , 1 ] 2 + 3 2 1 + 1 + 4 × [ 1 , 1 , 2 ] = [ 2 , 3 , 1 ] 3 6 × [ 1 , 1 , 2 ] Projection = [ 2 , 3 , 1 ] 2 + 3 2 1 + 1 + 4 × [ 1 , 1 , 2 ] = [ 2 , 3 , 1 ] 3 6 × [ 1 , 1 , 2 ] “Projection”=[2,3,-1]-(2+3-2)/(1+1+4)xx[1,1,2]=[2,3,-1]-(3)/(6)xx[1,1,2]\text{Projection} = [2, 3, -1] – \frac{2 + 3 – 2}{1 + 1 + 4} \times [1, 1, 2] = [2, 3, -1] – \frac{3}{6} \times [1, 1, 2]Projection=[2,3,1]2+321+1+4×[1,1,2]=[2,3,1]36×[1,1,2]
Projection = [ 2 , 3 , 1 ] [ 0.5 , 0.5 , 1 ] = [ 1.5 , 2.5 , 2 ] Projection = [ 2 , 3 , 1 ] [ 0.5 , 0.5 , 1 ] = [ 1.5 , 2.5 , 2 ] “Projection”=[2,3,-1]-[0.5,0.5,1]=[1.5,2.5,-2]\text{Projection} = [2, 3, -1] – [0.5, 0.5, 1] = [1.5, 2.5, -2]Projection=[2,3,1][0.5,0.5,1]=[1.5,2.5,2]

Step 4: Find the Equation of the Projected Line

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 ) (1,1,-1)(1, 1, -1)(1,1,1) on the original line, the equation of the projected line is:
x 1 1.5 = y 1 2.5 = z + 1 2 x 1 1.5 = y 1 2.5 = z + 1 2 (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}x11.5=y12.5=z+12

Conclusion

The projection of the given straight line x 1 2 = y 1 3 = z + 1 1 x 1 2 = y 1 3 = z + 1 1 (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)\frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1}x12=y13=z+11 onto the plane x + y + 2 z = 6 x + y + 2 z = 6 x+y+2z=6x+y+2z=6x+y+2z=6 is the line x 1 1.5 = y 1 2.5 = z + 1 2 . x 1 1.5 = y 1 2.5 = z + 1 2 . (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}.x11.5=y12.5=z+12.
Verified Answer
5/5
c^2=a^2+b^2-2ab\:Cos\left(C\right)
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