# Vector of triangle height constructed over two vectors

Given vectors: $\overrightarrow{a}=\overrightarrow{p}+2\overrightarrow{q},\overrightarrow{b}=3\overrightarrow{p}-\overrightarrow{q}$ where $|\overrightarrow{p}|=2,|\overrightarrow{q}|=6,\angle(\overrightarrow{p},\overrightarrow{q})=\pi/3$. Find vector of triangle height constructed over vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ that is orthogonal to vector $\overrightarrow{a}$.

Let the vector of triangle height is $\overrightarrow{h}$. Since $\overrightarrow{h}$ is orthogonal to $\overrightarrow{a}$, we can look at right-angled triangle formed of vectors $\overrightarrow{h},\overrightarrow{x_1},\overrightarrow{x}$ where $\overrightarrow{x_1}$ is the component of $\overrightarrow{x}$ onto $\overrightarrow{a}$ and $\overrightarrow{x}$ is hypotenuse.

Orthogonal projection of $\overrightarrow{x}$ onto $\overrightarrow{a}$ is given by $$\overrightarrow{x_1}=proj_{\overrightarrow{a}} {\overrightarrow{x}}=\frac{\overrightarrow{a}\cdot \overrightarrow{x}}{|\overrightarrow{a}|^2}\cdot \overrightarrow{a}$$

But we don't know the coordinates or magnitude of $\overrightarrow{a}$ and $\overrightarrow{x}$.

Question: How to find vector $\overrightarrow{h}$?

Dot product of $\overrightarrow{p},\overrightarrow{q}$ is $|\overrightarrow{p}\cdot \overrightarrow{q}|=|\overrightarrow{p}||\overrightarrow{q}|\cos\angle(\overrightarrow{p},\overrightarrow{q})=6$.

Cross product of $\overrightarrow{p},\overrightarrow{q}$ is $|\overrightarrow{p}\times \overrightarrow{q}|=|\overrightarrow{p}||\overrightarrow{q}|\sin\angle(\overrightarrow{p},\overrightarrow{q})=6\sqrt 3$.

Cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is $|\overrightarrow{a}\times \overrightarrow{b}|=6(6\sqrt 3-11)$.

HINTS:

You write of vector $\vec x$, but you really mean vector $\vec b$, at least if the question (which is badly worded) makes any sense.

You have enough information to find $|\vec a|$, $|\vec b|$, and $\vec a\cdot\vec b$. For example,

\begin{align} |\vec a| &= \sqrt{\vec a\cdot\vec a} \\[2ex] &= \sqrt{(\vec p+2\vec q)\cdot(\vec p+2\vec q)} \\[2ex] &= \sqrt{\vec p\cdot\vec p+4\vec p\cdot\vec q+4\vec q\cdot\vec q} \\[2ex] &= \sqrt{|\vec p|^2+4(\vec p\cdot\vec q)+4|\vec q|^2} \\[2ex] &= \sqrt{2^2+4(6)+4(6^2)} \\[2ex] &= \sqrt{172} \\[2ex] &= 2\sqrt{43} \end{align}

Then use those values to find $\vec{x_1}$ and thus $\vec h$.

• Dot product $\overrightarrow{a}\cdot \overrightarrow{b}$ gives negative result $-30$. Does this mean there is a mistake in the question? – user300048 Mar 28 '16 at 14:30
• @user300044: No, that dot product is correct. The angle between $\vec a$ and $\vec b$ is obtuse. You can check your answers, as I did, with Geogebra, by setting $p=(2,0),\ q=(3,3\sqrt 3)$. – Rory Daulton Mar 28 '16 at 14:33
• Vector of trisngle height is $\overrightarrow{h}=\overrightarrow{b}-\overrightarrow{x_1}=273/86 \overrightarrow{p}-28/43 \overrightarrow{q}$. – user300048 Mar 28 '16 at 14:59
• @user300044: Yes, that is correct! Again, you would be wise to check this with something like Geogebra, as I did. Also, this assumes my understanding of what the question actually asks is correct. You have probably seen that the "altitude", $\vec h$, is not actually inside the triangle formed by $\vec a$ and $\vec b$. – Rory Daulton Mar 28 '16 at 16:29
• @user300044: If my answer was helpful, please upvote and/or accept it, though you may want to wait longer for other answers. – Rory Daulton Mar 28 '16 at 16:33