Question #247163

Let L be the line given by the vector equation (x,y)=(1,1) + t( sqrt(3),1), tϵ\epsilonR. What is the equation of the image of L after being rotated 15 deg about (1,1) and then translated by vector u=(-1,1)


1
Expert's answer
2022-01-31T09:43:12-0500

The equation of a line in vector form is a+λba+\lambda b

where a is the position vector of a point on the line and b is a vector parallel to the line.

Rotating the line through and angle θ\theta is equivalent to rotating b through the same angle.

In this case: b=(3,1)b=(\sqrt 3,1)

The slope of the line segment from the origin to the head of this vector is 1/31/\sqrt 3

and hence its inclination to the positive X axis is arctan(1/3)arctan(1/\sqrt 3)

If we rotate this vector by 15°15\degreeits inclination to the positive X axis becomes

arctan(1/3)+15°arctan(1/\sqrt 3)+15\degree

Hence the slope of b1 after rotating:

tan(arctan(1/3)+15°)=tan45°=1tan(arctan(1/\sqrt 3)+15\degree)=tan45\degree =1

b1 can then be represented by the position vector of point (1,1)(1,1)


Translating the line through u=(-1,1) is equivalent to translating the given point on the line by

(1,1)(-1,1)

Hence, after this translation the point (1,1) becomes (0,2)

Thus, it can be seen that after the given rotation and translation the vector equation of the line becomes

(x,y)=(0,2)+t(1,1)(x,y)=(0,2)+t(1,1)


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