0
$\begingroup$

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 $|\overrightarrow{a}\times \overrightarrow{b}|=6(6\sqrt 3-11)$.

$\endgroup$
2
$\begingroup$

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 $\vec h$.

$\endgroup$
  • $\begingroup$ Dot product $\overrightarrow{a}\cdot \overrightarrow{b}$ gives negative result $-30$. Does this mean there is a mistake in the question? $\endgroup$ – user300048 Mar 28 '16 at 14:30
  • $\begingroup$ @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 $p=(2,0),\ q=(3,3\sqrt 3)$. $\endgroup$ – Rory Daulton Mar 28 '16 at 14:33
  • $\begingroup$ Vector of trisngle height is $\overrightarrow{h}=\overrightarrow{b}-\overrightarrow{x_1}=273/86 \overrightarrow{p}-28/43 \overrightarrow{q}$. $\endgroup$ – user300048 Mar 28 '16 at 14:59
  • $\begingroup$ @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 $\vec b$. $\endgroup$ – Rory Daulton Mar 28 '16 at 16:29
  • $\begingroup$ @user300044: If my answer was helpful, please upvote and/or accept it, though you may want to wait longer for other answers. $\endgroup$ – Rory Daulton Mar 28 '16 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.