# How do I determine a vector given angle and magnitude?

Question: Determine the vector $$d$$ that is perpendicular to $$c = 4i-3j$$ and has a magnitude of $$10$$.

My workings: Using the dot product:

First, I said vector $$d = xi +yj.$$

Since $$c$$ and $$d$$ are perpendicular to each other, the angle between them is $$90^o$$; cos($$90^o)=0$$ and therefore $$d\cdot c= 0.$$

$$d\cdot c$$ is also = $$(4 \times x)$$ + $$(-3 \times y)= 4x-3y.$$

Then $$|d|= 10$$; $$|d|$$ is also equal to $$\sqrt{x^2 +y^2}.$$

I then tried solving for $$x$$ and $$y$$ using simultaneous equations and it does not work (please explain); also please show workings. Textbook answer: $$\pm (6i+8j)$$.

You were on the right track.

To solve $$4x-3y=0$$ and $$x^2+y^2=10^2:$$

from the first equation $$x=\frac 3 4 y$$;

plug that expression for $$x$$ into the second equation and solve for $$y$$.

• wait, does $\sqrt{x^2 +y^2}$ = $x +y$ ? I tried like 50 times i could not do it. – Nhoj_Gonk Mar 5 '19 at 12:35
• @FredWeasley: no, not in general – J. W. Tanner Mar 5 '19 at 12:36
• So i cannot take root of both sides for $x^2 +y^2$ = $10^2$ ? – Nhoj_Gonk Mar 5 '19 at 12:38
• @FredWeasley: that mistake is called Freshman's dream; I can disprove it with a counterexample: $\sqrt{6^2+8^2}=10$ but $6+8=14$ – J. W. Tanner Mar 5 '19 at 12:51
• If $\sqrt{x^2+y^2}=x+y$ then $x^2+y^2=(x+y)^2=x^2+y^2+2xy,\,$ so $2xy=0,\,$ so $x=0$ and/or $y=0$ – J. W. Tanner Mar 5 '19 at 12:59