tearstreakdl

2021-12-14

Find the values of x such that the angle between the vectors (2, 1, -1), and (1, x, 0) is ${45}^{\circ }$.

Hector Roberts

Theorem 3 from this section states that:
$a\cdot b=midaparal\le lbmid\mathrm{cos}\theta$
where \theta is the angle between the vectors a,b
$⟨2,1,-1⟩\cdot ⟨1,x,0⟩=\sqrt{{2}^{2}+{1}^{2}+{\left(-1\right)}^{2}}\sqrt{{1}^{2}+{x}^{2}+{0}^{\left\{}}{\mathrm{cos}45}^{\circ }$
Remember that:
$⟨{a}_{1},{b}_{1},{c}_{1}⟩\cdot ⟨{a}_{2},{b}_{2},{c}_{2}⟩={a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}$
This is due to the fact that the unit vectors I j, and k have a dot product of 1 with themselves and a dot product of 0 with the other two.
$\left(2\right)\left(1\right)+\left(1\right)\left(x\right)+\left(-1\right)\left(0\right)=\sqrt{4+1+1}\sqrt{1+{x}^{2}+0}\frac{1}{\sqrt{2}}$
$2+x+0=\sqrt{6}\sqrt{1+{x}^{2}}\frac{1}{\sqrt{2}}$
$2+x=\sqrt{3}\sqrt{1+{x}^{2}}$
Square both sides
${\left(2+x\right)}^{2}=3\left(1+{x}^{2}\right)$
$4+4x+{x}^{2}=3+3{x}^{2}$
The quadratic equation should now be written in standard form.
$2{x}^{2}-4x-1=0$
Using the quadratic formula, we can write
$x=\frac{-\left(-4\right)±\sqrt{{\left(-4\right)}^{2}-4\left(2\right)\left(-1\right)}}{2\left(2\right)}=\frac{4±\sqrt{24}}{2\left(2\right)}=\frac{2±\sqrt{6}}{2}$

sukljama2

Step 1
Consider the vectors:
$v=⟨2,1,-1⟩$
$w=⟨1,x,0⟩$
$\theta ={45}^{\circ }$
$\stackrel{\to }{v}2i+j-k$
$\stackrel{\to }{w}=i+xj$
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=\left(2i+j-k\right)\cdot \left(i+k\right)$
$=\left(2\right)\left(1\right)+\left(1\right)\left(x\right)+\left(-1\right)\left(0\right)$
$=2+x$
Step 2
Find the magnitude each vector:
$mid\stackrel{\to }{v}mid=\sqrt{{2}^{2}+{1}^{2}+{\left(1\right)}^{2}}$
$=\sqrt{4+1+1}$
$=\sqrt{6}$
$mid\stackrel{\to }{w}mid=\sqrt{{1}^{2}+{\left(x\right)}^{2}+{\left(0\right)}^{2}}$
$=\sqrt{1+{x}^{2}+0}$
$=\sqrt{1+{x}^{2}}$
Step 3 Find the cosine angle between two vectors:
$\stackrel{\to }{v}\cdot \stackrel{\to }{w}=mid\stackrel{\to }{v}paral\le l\stackrel{\to }{w}mid\mathrm{cos}\theta$
$2+x=\left(\sqrt{6}\right)\left(\sqrt{1+{x}^{2}}\right\}{\mathrm{cos}45}^{\circ }$
$2+x=\left(\sqrt{6}\right)\left(\sqrt{1+{x}^{2}\right)}\left(\begin{array}{c}\frac{1}{\sqrt{2}}\end{array}\right)$
$2+x=\left(\sqrt{6}\right)\left(\sqrt{1+{x}^{2}\right)}\left(\begin{array}{c}\frac{\sqrt{2}}{2}\end{array}\right)$
$2+x=\frac{\sqrt{12}}{2}\left(\sqrt{1+{x}^{2}}\right\}$
$\frac{\sqrt{12}}{2}\left(2+x\right)=\sqrt{1+{x}^{2}}$
$\frac{4}{12}{\left(2+x\right)}^{2}=1+{x}^{2}$
$\frac{4}{12}\left(4+{x}^{2}+4x\right)=1+{x}^{2}$
$\frac{1}{3}\left(4+{x}^{2}+4x\right)=1+{x}^{2}$

Do you have a similar question?