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 $y$.

  • $\begingroup$ wait, does $\sqrt{x^2 +y^2}$ = $x +y$ ? I tried like 50 times i could not do it. $\endgroup$ – Nhoj_Gonk Mar 5 '19 at 12:35
  • $\begingroup$ @FredWeasley: no, not in general $\endgroup$ – J. W. Tanner Mar 5 '19 at 12:36
  • $\begingroup$ So i cannot take root of both sides for $x^2 +y^2$ = $10^2$ ? $\endgroup$ – Nhoj_Gonk Mar 5 '19 at 12:38
  • 1
    $\begingroup$ @FredWeasley: that mistake is called Freshman's dream; I can disprove it with a counterexample: $\sqrt{6^2+8^2}=10$ but $6+8=14$ $\endgroup$ – J. W. Tanner Mar 5 '19 at 12:51
  • 1
    $\begingroup$ 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 $y=0$ $\endgroup$ – J. W. Tanner Mar 5 '19 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.