Angle o formed between the two vectors? A. ucdot v=vert uVert vvert sinTheta C. ucdot v=-vert uVert vvert cosTheta B. ucdot v=-vert uVert vvert sinTheta D ucdot v=vert uVert vvert cosTheta 32. A translation takes a point (1,2) to (3,3) Then what is the equation of the image of the circle whose equation x^2+y^2=25 A. (x+2)^2+(y+1)^2=25 B (x-1)^2+(y-2)^2=25 C. (x-2)^2+(z-1)^2=25 D. x^2+y^2=25 33. Let the point P=(4,6) be reflected about the line L: y=-3x+2 Then what is the image of the point P? A. ((14)/(5),-(7)/(5)) C. ((23)/(5),-(14)/(5)) (-(28)/(5),(14)/(5)) D. (-(14)/(5),(7)/(5)) 34. Let a point (x,y) be rotated counter-clockwise direction through the angle Theta =(pi )/(2) Then what

Let's correct and complete the answers for the given questions:<br /><br />**Question 1: Angle θ formed between two vectors**<br /><br />The correct answer is:<br />D. \( u \cdot v = |u| |v| \cos(\theta) \)<br /><br />This formula represents the dot product of two vectors \( u \) and \( v \), which is equal to the product of their magnitudes and the cosine of the angle between them.<br /><br />**Question 2: Translation of a circle**<br /><br />Given the translation takes a point \((1,2)\) to \((3,3)\), the translation vector is \((2,1)\). To find the equation of the image of the circle \( x^2 + y^2 = 25 \) after translation, we need to shift the center of the circle by the translation vector.<br /><br />The original center of the circle is \((0,0)\). After translation, the new center is \((0+2, 0+1) = (2,1)\).<br /><br />So, the equation of the translated circle is:<br />B. \( (x-2)^2 + (y-1)^2 = 25 \)<br /><br />**Question 3: Reflection of a point about a line**<br /><br />Given the point \( P = (4,6) \) and the line \( L: y = -3x + 2 \), we need to find the image of the point after reflection.<br /><br />First, find the perpendicular distance from the point to the line. The formula for the perpendicular distance from a point \((x_1, y_1)\) to a line \( Ax + By + C = 0 \:<br />\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]<br /><br />For the line \( y = -3x + 2 \), rewrite it as \( 3x + y - 2 = 0 \). Here, \( A = 3 \), \( B = 1 \), and \( C = -2 \).<br /><br />\[ d = \frac{|3(4) + 1(6) - 2|}{\sqrt{3^2 + 1^2}} = \frac{|12 + 6 - 2|}{\sqrt{9 + 1}} = \frac{16}{\sqrt{10}} = \frac{8\sqrt{10}}{5} \]<br /><br />The reflection point will be at a distance \( d \) on the other side of the line. The slope of the line is \(-3\), so the slope of the perpendicular line is \(\frac{1}{3}\).<br /><br />Using the point-slope form of the line equation:<br />\[ y - 6 = \frac{1}{3}(x - 4) \]<br /><br />Solving for the reflection point:<br />\[ y = -3x + 2 \]<br />\[ y - 6 = \frac{1}{3}(x - 4) \]<br />\[ 3(y - 6) = x - 4 \]<br />\[ 3y - 18 = x - 4 \]<br />\[ x =y - 14 \]<br /><br />Substitute \( y = -3x + 2 \):<br />\[ x = 3(-3x + 2) - 14 \]<br />\[ x = -9x + 6 - 14 \]<br />\[ 10x = -8 \]<br />\[ x = -\frac{8}{10} = -\frac{4}{5} \]<br /><br />Then, \( y = -3(-\frac{4}{5}) + 2 = \frac{12}{5} + 2 = \frac{22}{5} \).<br /><br />So, the reflected point is:<br />D. \( \left(-\frac{14}{5}, \frac{7}{5}\right) \)<br /><br />**Question 4 of a point**<br /><br />When a point \((x, y)\) is rotated counterclockwise by an angle \(\Theta = \frac{\pi}{2}\), the new coordinates are:<br />\[ (x', y') = (-y, x) \]<br /><br />So, if the original point is \((x, y)\), the new point after rotation is:<br />\[ (x', y') = (-y, x) \]<br /><br />This completes the answers for the given questions.