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I urgently need help correcting a multivariable calculus exam involving vectors. The questions are listed below, along with the answers that I put:

1) Let $P= (1, 0, 1)$, $Q = (4, 1, 9)$, and $R= (3, 2, 8)$ Calculate the area of the triangle whose vertices are $P$, $Q$ and $R$.

What I did for this question was form vectors ${\bf PQ}= <3, 1, 8>$ and ${\bf PR}= <2, 2, 7>$, and took the cross product of the vectors to give me $(122/2)^{1/2}$ units squared. The teacher still counted the question wrong however.

2) A 5 ounce swallow is flying through the land of America carrying a 1 pound apple by a 1/2 ounce strand of tree bark held under one of its dorsal guiding feathers. Along the way out intrepid swallow stops to rest on a 73 in tree limb that makes an angle of $112 ^\circ$ with the trunk of the tree. Find the component forms of the limb and bird vectors and use a vector product to calculate the torque, in pythons (a made up unit), on the limb caused by the swallow and his cargo. 16 ounces are in a pound.

What I am confused about in the questions was finding the $z$ component for the bird's vector. I was not sure if it is $73 \sin(22 ^\circ)-6.5 \text{ oz}$, taking into account the initial height of the tree limb and how high up the bird was upon landing on the tree limb. I also need help finding the z component of the tree limb. I just need to be walked through these parts; I will be fine calculating the torque.

3) Let u$ = <5, 3, -2>$ and v$ = <6, 3, -12> $ Decompose u into a sum of two vectors pp and w, where p is parallel to v, and w is orthogonal to v.

I took the cross product of u and v to find a vector that could make w when multiplied by the right scalar, but I do not know what said scalar is. On the other hand, I was hoping to solve for p by subtracting w from u, and just writing a scalar variable next to the p that I obtained that would be distributed to the components of p.

I think I have corrected the following question, but I would like to have my work checked.

4) still letting p= <5, 3, -2> and v= <6, 3, -12>, Suppose F1 is a force vector in the direction of u with a magnitude of 308 decalopans (dal) and F2 is a force vector in the direction of v with a magnitude of 227 dal. Find the magnitude and direction of the resultant force F= F1+F2.

To find the resultant force's magnitude, I just added the magnitudes of F1 and F2, and got 535 dal. To find the direction of the resultant force, I just added u and v together to get new vector. I then added all of that components together, divided the sum by said vector's magnitude, and took the inverse cosine of the resulting quotient to get the direction in theta.

In other words, I did u+v and got $<11, 6, -14>$. I added these components together and got 3, which I divided by the magnitude, with would be $353^{1/2}$. The cosine inverse of the quotient was $80.81 ^\circ$.

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  • $\begingroup$ An advice, which might be beneficiary to you: meta.stackexchange.com/questions/39223/… $\endgroup$ – eudes Feb 7 '19 at 22:53
  • $\begingroup$ In 1), should it be "calculate the area of the triangle" and (your result) "$(122/2)^{\wedge} (1/2)$"? Because to me, to calculate a triangle means to find its dimensions (eg. all sides, or a side and the angles). I'll edit, you can accept if yes. $\endgroup$ – eudes Feb 7 '19 at 23:16
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1) By "calculate the triangle", do you mean "calculate the area of the triangle"?
In any case, the area is half the length of the cross product. You found $122$, the square of the length of the cross product - and then squared it instead of taking a square root. It should have been $\frac{\sqrt{122}}{2}$.

2) All of the forces exerted on the limb are purely vertical - the weight of the limb, the weight of the bird and cargo, and the counterweight force of the connection to the trunk holding it up. So, if you're using the standard coordinates, the $z$ coordinate of the vector for the bird's force is simply the weight (directed downwards). The $z$ coordinate of the vector for the difference in position between the bird and the connection to the trunk is the length times that $\sin(22^\circ)$ as you had. Upwards or downwards? I don't know - the problem setup isn't clear on that. (This also uses the unstated assumptions that the trunk is vertical and that the bird landed at the end. We measure torque at the connection point, where it's most natural and we don't have to worry about how the counterweight force changes.)
On the other hand, the $z$ coordinate of the trunk vector doesn't really matter. Torque is a cross product; that coordinate is just going to get multiplied by the zeros in the horizontal directions for the bird's weight.

3) No, the cross product isn't what you want here. The cross product of $u$ and $v$ is a vector orthogonal to both $u$ and $v$, while the $w$ you seek is in the span of $u$ and $v$ - it will be orthogonal to $u\times v$ and never a constant multiple of it. What you're looking for is the orthogonal projection of $u$ onto $v$ to find $p$, and then subtracting that from $p$ to get $w$.

4) The magnitude of the sum of two vectors is only equal to the sum of the magnitudes if those vectors are parallel. These aren't. We have to set up a sum of vectors by putting $F_1$ and $F_2$ in component form. How? For $u$, that would be the magnitude $308$ times a unit vector $\frac{\langle 5,3,-2\rangle}{\sqrt{5^2+3^2+2^2}}$ in the $u$ direction. $F_2$ is similar.

I added these components together and got 3, which I divided by the magnitude, with would be (353)^(1/2). The cosine inverse of the quotient was 80.81 degress.

This is not a procedure that makes sense. For one, it's not guaranteed to produce a valid angle; if we applied it to the vector $\langle 1,1,1\rangle$, we would get $\sqrt{3}$, which doesn't have an arccosine. More importantly, it takes more than one angle to specify a direction in $\mathbb{R}^3$. The longitude and colatitude angles from spherical coordinates would work, but usually we just specify the direction with a unit vector in that direction.

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  • $\begingroup$ Thank you so much jmerry. I actually made a typo on the first question, but I did have 1/2 (122)^(1/2) units squared (units squared being my units). $\endgroup$ – Uchuuko Feb 8 '19 at 3:11
  • $\begingroup$ Yes, I did mean "calculate the area of the triangle". $\endgroup$ – Uchuuko Feb 8 '19 at 4:09
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Re 1.
Firstly, the length of the cross product is actually the area of the parallelogram given by the two vectors, so the area of the triangle is half of that. I've got the length of the product as $\sqrt{122}$, so the number $122$ seems to be right, but something must be wrong later in your formulas. Then the area of the triangle is $\frac 1 2 \sqrt{122}$.

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  • $\begingroup$ Thank you eudes. I had a typo and meant to type your answer, but I had units squared as my units. $\endgroup$ – Uchuuko Feb 8 '19 at 3:12
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Re 3.
Let $\bf u = p + w$, as per your notation, and also $\mathbf p = a\mathbf v$. Then the cross product:
$$\bf u\times v = (p+w)\times v = w\times v, {\rm\text { since }} p\times v = {\mathit a} v\times v = 0.$$ Now $\bf w \perp v$, and $\bf w\times v$ is orthogonal to both, so $\bf v\times (w\times v)$ is parallel to $\bf w$ and has the same direction (explanation below). Now you only need to get the magnitude of $\bf w$ right. This is quite easy, as many of our vectors are pairwise orthogonal: $$ \begin{gather} \|\bf w\times v \| = \|w \| \, \|v\| \sin \angle(w,v) = \|w \| \, \|v\| \\ \bf \|w\| = \frac {\|w \times v\|}{\|v\|} = \frac {\|u \times v\|}{\|v\|} = \|u\|\sin\angle(u,v) \tag{!} \\ \bf\| v\times (w\times v)\| = \|v\| \, \|w\| \, \|v\| \\ \bf w = \frac { v\times (w\times v)}{\| v\times (w\times v)\|} \|w\| = \frac { v\times (u\times v)}{\|v\| \, \|w\| \, \|v\|} \|w\| \\ \bf w = \frac 1{\|v\|^2} v\times (u\times v) = \frac v{\|v\|} \times (u\times \frac v{\|v\|}) \end{gather}$$ However, there's an easier way. Because the scalar product of orthogonal vectors is zero: $$ \mathbf u\cdot \mathbf v = \mathbf p\cdot \mathbf v = a \mathbf v\cdot \mathbf v = a \|\mathbf v\|^2, $$ whence $$ \mathbf p = a\bf v = \frac {u\cdot v}{\| v\|^2} v = \bigg(u\cdot \frac{v}{\| v\|}\bigg) \frac v{\| v\|} $$ $$ \bf \|p\| = \|u\| \cos\angle(u,\frac v {\|v\|}) = \|u\| \cos\angle(u,v). \tag{!} $$ Finally $$ \bf u = \frac 1{\|v\|^2} \bigg( (u\cdot v) \,v + v\times (u\times v)\bigg) $$ and if $\bf v$ is normalized: $$ \bf u = (u\cdot v)\,v + v\times (u\times v) = (v\cdot u)\,v + (v\times u)\times v. $$


Why are $\bf v\times(w\times v)$ and $\bf w$ of the same direction.
Imagine how those vectors are situated: $\bf w$ is orthogonal to $\bf v$ and $\bf w\times v$, and so is $\bf v\times (w\times v)$, thus they are parallel (all vectors that are simultaneously orthogonal to two nonparallel vectors, form a line in $\mathbb{R}^3$, so all are parallel). They are of the same direction, because any two orthogonal vectors and their product form a positively oriented triple (by definition of the cross product), and so do cyclic permutations of them: $$ \bf w, \ v, \ w\times v \quad \longrightarrow \quad v, \ w\times v, \ w \quad \longrightarrow \quad w\times v, \ w, \ v. $$ The triple $\bf v, \ w\times v, \ v\times (w \times v)$ is also positively oriented (again, two vectors and their cross product), so same as $\bf v, \ w\times v, \ w$.

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  • $\begingroup$ Thank you for the thorough explanation! :) $\endgroup$ – Uchuuko Feb 8 '19 at 3:13
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The first question can be done by 1/2 the determinant of the coordinates.

$$\frac{1}{2} \begin{Vmatrix} 1&0&1\\ 4&1&9\\ 3&2&8 \end{Vmatrix} = \frac{15}{2}$$

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