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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: $\pm (6i+8j)$.

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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 $y$.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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

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