1
$\begingroup$

Determine the vector $\vec d$ perpendicular to the z axis and to the vector $\vec a$ = 8$\vec i$ − 15$\vec j$ + 3$\vec k$ , so that its intensity is 50 and forms an acute angle with the x axis.


So here is what I know:

$\vec{d} = (d_1, d_2, d_3) $

$ \vec{a} \cdot \vec{d} = 0 $

This translates into $ 8d_1 - 15d_2 + 3d_3 = 0 $

$ | \vec{d} | = 50 $ ->

$ \sqrt{d_1^2 + d_2^2 + d_3^2}=50 $

But I dont know what does it mean that vector $\vec d$ is perpendiuclar to the z axis, and that vector $\vec d$ forms an acute angle with x axis. I would be thankful for any help.

$\endgroup$
3
  • 2
    $\begingroup$ $\;d\;$ perpendicular to the $\;z\,-$ axis $\;\iff d\perp(0,0,1)\;$ $\endgroup$
    – DonAntonio
    Feb 24, 2023 at 20:20
  • 1
    $\begingroup$ As in your previous question, cosine of $\vec d$'s angle with the $x$-axis becomes $d_1$. If that angle is acute, then its cosine is positive. $\endgroup$
    – peterwhy
    Feb 24, 2023 at 20:31
  • $\begingroup$ Yes, I figure it out when I saw the first comment and I remmbered that it was the same as my prevoius question. Thank you @peterwhy $\endgroup$ Feb 24, 2023 at 20:45

0

You must log in to answer this question.

Browse other questions tagged .