# Find vector d perpendicular to the z axis and to the vector a

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.

• $\;d\;$ perpendicular to the $\;z\,-$ axis $\;\iff d\perp(0,0,1)\;$ Feb 24, 2023 at 20:20
• 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. Feb 24, 2023 at 20:31
• 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 Feb 24, 2023 at 20:45