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I have two vectors, $(-2, 3, 1)$, and $(-1, 2, a)$. I know that the angle between these two vectors is $40^\circ$. How do I find a?

When I try to algebraically solve for a, I run into problems with the magnitudes. I understand that $X \cdot Y = |X| |Y| cos(Z)$. The problem is the magnitude of $Y$ in this case turns out to be $\sqrt{5 + a^2}$, and the dot product of $X$ and $Y$ is a + 8, so I can't figure out how to resolve the two to find a.

$a + 8 = \sqrt{14} \cdot \sqrt{5 + a^2} \cdot \cos(40^\circ)$

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    $\begingroup$ Welcome to math.se. Here are some tips on how to ask a good question. Please include your thoughts on the problem. $\endgroup$
    – Viktor Vaughn
    Apr 19, 2021 at 22:18
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    $\begingroup$ Do you remember how angle between two vectors is defined? It has something to do with dot product. Now... describe the information you are given as a formula and algebraically solve for $a$. $\endgroup$
    – JMoravitz
    Apr 19, 2021 at 22:18
  • $\begingroup$ When I try to algebraically solve for a, I run into problems with the magnitudes. I understand how X dot product Y = |X| * |Y| cos(Z). The problem is the magnitude of Y in this case turns out to be sqrt(5 + a^2), and the dot product of X and Y is a + 8, so I can't figure out how to resolve the two to find a. $\endgroup$
    – Joshua Like
    Apr 19, 2021 at 22:52
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    $\begingroup$ That looks like a reasonable equation to solve for $a$. $\endgroup$
    – Tunococ
    Apr 19, 2021 at 23:05
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    $\begingroup$ @JoshuaLike You might need to use the general formula for the quadratic equation (en.wikipedia.org/wiki/Quadratic_formula) to solve for $a$. $\endgroup$
    – Tunococ
    Apr 19, 2021 at 23:13

1 Answer 1

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We know that the scalar product is defined as:

$$\mathbf{v}\bullet \mathbf{u}=vu\cos \theta\equiv v_xu_x+v_yu_y+v_zu_z$$

i.e. $$\sqrt{14}\cdot \sqrt{5+a^2}=\frac{8+a}{\cos(40^{\circ})} \iff \sqrt{14\cdot (5+a^2)}=\frac{8+a}{\cos(40^{\circ})}$$

Hence, squaring LHS and RHS,

$$70\cos ^2\left(40^{\circ}\right)+14\cos ^2\left(40^{\circ }\right)a^2=64+16a+a^2$$

and the solution are (equation of second degree or quadratic formula), being the $70+14a^2>0, \, \forall a\in \Bbb R$,

$$a_1=\frac{-16+\sqrt{-3920\cos ^4\left(40^{\circ}\right)+3864\cos ^2\left(40^{\circ}\right)}}{2\left(1-14\cos ^2\left(40^{\circ}\right)\right)}, $$ $$a_2=-\frac{\sqrt{-3920\cos ^4\left(40^{\circ}\right)+3864\cos ^2\left(40^{\circ}\right)}+16}{2\left(1-14\cos ^2\left(40^{\circ }\right)\right)}$$

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    $\begingroup$ I am at this point myself, but I cannot separate a in order to find the possible values. Is there a way to do this that I am missing? $\endgroup$
    – Joshua Like
    Apr 19, 2021 at 23:10
  • $\begingroup$ @JoshuaLike I have finished the various edit. I hope that the steps are clear. $\endgroup$
    – Sebastiano
    Apr 19, 2021 at 23:25
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    $\begingroup$ Thank you @Sebastiano! This was very helpful! $\endgroup$
    – Joshua Like
    Apr 19, 2021 at 23:51
  • $\begingroup$ @JoshuaLike I am glady to help you. My best regards. $\endgroup$
    – Sebastiano
    Apr 20, 2021 at 9:53

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