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

[Niamh1]

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

 

[Evgeny.Makarov]

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.

 

[Prove It]

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:

$\displaystyle \begin{align*} \left| \mathbf{a} - \mathbf{b} \right| ^2 &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a}\right| \left| \mathbf{b}\right| \, \cos{ \left( \theta \right) } \\ \left( a_1 - b_1 \right) ^2 + \left( a_2 - b_2 \right) ^2 + \left( a_3 - b_3 \right) ^2 &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ a_1^2 -2\,a_1\,b_1 + b_1^2 + a_2^2 - 2\,a_2\,b_2 + b_2^2 + a_3^2 -2\,a_3\,b_3 + b_3^2 &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 - 2\,\left( a_1\,b_1 + a_2\,b_2 + a_3\,b_3 \right) &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\mathbf{a}\cdot \mathbf{b} &= \left| \mathbf{a} \right| ^2 + \left| \mathbf{b} \right| ^2 - 2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ -2\,\mathbf{a}\cdot \mathbf{b} &= -2\,\left| \mathbf{a} \right| \left| \mathbf{b} \right| \,\cos{ \left( \theta \right) } \\ \mathbf{a}\cdot \mathbf{b} &= \left| \mathbf{a} \right| \left| \mathbf{b} \right|\,\cos{ \left( \theta \right) } \\ \cos{ \left( \theta \right) } &= \frac{\mathbf{a}\cdot \mathbf{b}}{\left| \mathbf{a} \right| \left| \mathbf{b} \right| } \end{align*}$