Question

Find a vector of length 3 that is perpendicular to [4, 4, 7]

Answer 1

\[4a+4b+7c=0\] and \[a ^{2}+b ^{2}+c ^{2}=9\]

Answer 2

but we are one equation short

Answer 3

Yes I got that far, I've tried using guess and check but couldn't get any correct answers.

Answer 4

Looks like we have infinitely many possibilities

Answer 5

I can't figure out how to find one with length 3 though.

Answer 6

As long as we can have a vector such that when we take the dot product with it and your vector we get 0, that works for perpendicular. We can invent anything really. So for a dot product, we know we multiply the i, j, and k components and then add them together.

So we take 4(i) + 4(j) + 7(k) = 0 and choose an i, j, k that will make this true. Something simple that works is i = 1, j = -1, k = 0, which gives us the vector (1, -1, 0).

Now we just need this to be a length of 3. To do this, I'll turn this vector into a unit vector, which has a length of 1, and multiply it by 3. Doing this will make this vector into something of length 3 without changing its direction. So to get a unit vector out of this, we'll divide each component by the magnitude. So the magnitde is:
\[\sqrt{(1)^{2} + (-1)^{2} + (0)^{2}}= \sqrt{2}\] So our unit vector is then:
\[(\frac{ 1 }{ \sqrt{2} }, \frac{ -1 }{ \sqrt{2} },0)= (\frac{ \sqrt{2} }{ 2 },\frac{ -\sqrt{2} }{ 2 },0)\]So now we just multiply this by 3 and we're done, so your vector is:
\[(\frac{ 3\sqrt{2} }{ 2 },\frac{ -3\sqrt{2} }{ 2 },0)\] If you like, you could check that this vector is still perpendicular by reconfirming that the dot product is 0 and check to see that its magnitude is 3, but this would be an appropriate vector for you problem :)

Answer 7

Thank you so much!!