MHB Find value of x and area of parallelogram: a,b,c,P

AI Thread Summary
To find the value of x in vector a = (2, x, 0) such that the angle between vectors a and b = (1, 0, -1) is π/4, the cosine formula for angles between vectors is applied. The dot product of a and b, along with their magnitudes, is used to establish the relationship needed to solve for x. Once x is determined, the area of the parallelogram formed by the unit vector in the direction of b and vector c = (5, -9, 3) can be calculated using the cross product. The area is given by the magnitude of the vector product of the two vectors. This approach effectively combines vector algebra with geometric principles to solve the problem.
Niamh1
Messages
4
Reaction score
0
Hello, could someone please help me with this question? I don't even know where to begin.

Given vectors a = (2, x, 0), b = (1, 0, −1) and c = (5, −9, 3), and let P(2, 1, −1)
be a point. Find the value of x in a such that the angle between a and b is π/4, then find the area of parallelogram with adjacent sides are bˆ and c, where bˆ is the unit vector in the direction of b
 
Mathematics news on Phys.org
You should use the following facts.

The cosine of angle between $$v_1=(x_1,y_1,z_1)$$ and $v_2=(x_2,y_2,z_2)$ is $\dfrac{v_1\cdot v_2}{|v_1|\cdot|v_2|}$. Here $v_1\cdot v_2=x_1x_2+y_1y_2+z_1z_2$ is the dot product, or scalar product, of $v_1$ and $v_2$. The length of $v_1$ is denoted by $|v_1|$ and equals $\sqrt{v_1\cdot v_1}$.

The unit vector in the direction of $v$ is $\dfrac{v}{|v|}$.

The signed area of the parallelogram with adjacent sides $$v_1=(x_1,y_1,z_1)$$ and $v_2=(x_2,y_2,z_2)$ is $|v_1\times v_2|$. Here $v_1\times v_2$ is the vector product of $v_1$ and $v_2$, and it equals $\begin{vmatrix}i&j&k\\x_1&y_1&z_1\\x_2&y_2&z_2\end{vmatrix}$, where $i$, $j$ and $k$ are mutually perpendicular unit vectors in the directions of the coordinate axes.
 
Evgeny.Makarov said:
You should use the following facts.

The cosine of angle between $$v_1=(x_1,y_1,z_1)$$ and $v_2=(x_2,y_2,z_2)$ is $\dfrac{v_1\cdot v_2}{|v_1|\cdot|v_2|}$. Here $v_1\cdot v_2=x_1x_2+y_1y_2+z_1z_2$ is the dot product, or scalar product, of $v_1$ and $v_2$. The length of $v_1$ is denoted by $|v_1|$ and equals $\sqrt{v_1\cdot v_1}$.

The unit vector in the direction of $v$ is $\dfrac{v}{|v|}$.

The signed area of the parallelogram with adjacent sides $$v_1=(x_1,y_1,z_1)$$ and $v_2=(x_2,y_2,z_2)$ is $|v_1\times v_2|$. Here $v_1\times v_2$ is the vector product of $v_1$ and $v_2$, and it equals $\begin{vmatrix}i&j&k\\x_1&y_1&z_1\\x_2&y_2&z_2\end{vmatrix}$, where $i$, $j$ and $k$ are mutually perpendicular unit vectors in the directions of the coordinate axes.

Because the rule of the cosine of the angle is not entirely obvious, if you have two vectors $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$, then they can form two sides of a triangle, with the third side being the vector $\displaystyle \begin{align*} \mathbf{a} - \mathbf{b} \end{align*}$. Then if $\displaystyle \begin{align*} \theta \end{align*}$ is the angle between vectors $\displaystyle \begin{align*} \mathbf{a} \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} \end{align*}$, we can relate the four pieces of information with the cosine rule:

$\displaystyle \begin{align*} \left| \mathbf{a} - \mathbf{b} \right| ^2 &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a}\right| \left| \mathbf{b}\right| \, \cos{ \left( \theta \right) } \\ \left( a_1 - b_1 \right) ^2 + \left( a_2 - b_2 \right) ^2 + \left( a_3 - b_3 \right) ^2 &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ a_1^2 -2\,a_1\,b_1 + b_1^2 + a_2^2 - 2\,a_2\,b_2 + b_2^2 + a_3^2 -2\,a_3\,b_3 + b_3^2 &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 - 2\,\left( a_1\,b_1 + a_2\,b_2 + a_3\,b_3 \right) &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\mathbf{a}\cdot \mathbf{b} &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ -2\,\mathbf{a}\cdot \mathbf{b} &= -2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ \mathbf{a}\cdot \mathbf{b} &= \left| \mathbf{a} \right| \left| \mathbf{b} \right|\,\cos{ \left( \theta \right) } \\ \cos{ \left( \theta \right) } &= \frac{\mathbf{a}\cdot \mathbf{b}}{\left| \mathbf{a} \right| \left| \mathbf{b} \right| } \end{align*}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top