MTE1_A_Solution

# 2pt c

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Unformatted text preview: of p and q. a. Find the value(s) of the scalar product of p and q. [2pts] b. Suppose we change p and q so that the magnitudes of p and q remain 3 and 5 respectively but the angle between p and q is doubled. What is the scalar product of p and q? [2pt] c. Suppose p and q are changed as described in part b above. What is the magnitude of the cross product of p and q? [2pt] Answer: Let denote the angle between p and q. Then the magnitude of the scalar product is |15cos()| while the magnitude of the cross product is |15sin()|. We are given that these are equal. Therefore, |cos()|=|sin()|. Using elementary trigonometry, it must be the case that |cos()|=|sin()| = 1/ 2 . Therefore, possible values of are /4, 3/4, ‐ /4, ‐3/4. a. Scalar product of p and q is given by 15cos(). Possible values are 10.61 . b. Now, we are asked to double the angle while keeping magnitudes the same. Therefore, possible values of the angle are /2, 3/2, ‐ /2, ‐3/2 or just two values /2, ‐ /2. Scalar product of p and q then becomes 15cos(/2)=0=15cos(/2). Thus, the answer is zero. c. Magnitude of the cross product is |15sin(/2)|=15 in both cases. 3. Consider vectors x and y in 6 dimensions given by x [3 0 4 1 2 3] y [1 2 0 1 3 1] a. Find the component of vector x along vector y. [3pts] b. Define new vectors z=x+y and w=x‐y. Find the scalar product of z and w. [2pt] Answer: a. The formula for component of a vector x along vector y is (x.y)y/||y||. So what we need to do is to calculate x.y and ||y||. We get x.y=‐13 and ||y||=4. Therefore, the component of x along y is 13 / 4 13 / 2 0 13 / 4 39 / 4 13 / 4 For part b, note that z.w=(x+y).(x‐y)=x.x‐y.y=39‐16=23. 4. Let x an...
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## This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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