Calculus in Three Dimensions (21-259) — Feedback on Homework 2

Homework 2 was due on Tuesday 9th September 2014. Questions 1, 2, 3, 4 and 5 were graded.

Question 1. The most common error in part (a) was people simply writing the vector $\overrightarrow{AB}$ for the equation of the line. This is the direction of the line; the vector equation of the line is $\mathbf{r}(t) = \overrightarrow{OA} + t\overrightarrow{AB}$.

For part (b), the vector equation of a line segment is the same as the vector equation of a line, except the values of $t$ are restricted. The line segment from $A$ to $B$ is thus the same as in part (a), but with a restriction on $t$.

A common problem with part (c) was people taking the direction vector of the line as being the normal vector to the plane. This isn't true: the line lies in the plane, so in fact the plane is parallel to the direction vector of the line! You needed to use the fact that, say, both $\overrightarrow{AB}$ and $\overrightarrow{AC}$ lie in the plane, and take their cross product to find the normal vector.

Question 2. Most people who got this question wrong did so because they misunderstood what 'coordinate plane' meant. You had to find the values of $t$ for which $x=0$, $y=0$ or $z=0$, and then find the corresponding points on the line.

Question 3. As with 1(c), the main error here was taking the direction vector of the line as being the normal vector of the plane.

Question 4. This question was mostly done very well. Those who messed it up usually did so because they simply set the $x$, $y$ and $z$ coordinates equal to each other without renaming one of the parameters. The idea is: as the parameters vary, the vector traces out a line; the lines intersect, but the value of the parameters on the two lines at the point of intersection aren't the same.

Question 5. If you made it all the way to the end of this fiddly question then I commend you. It was done surprisingly well, with the most common error being people mixing up their $-$ signs when taking a cross product.

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