How do you find two unit vectors orthogonal to a=⟨−1,2,1⟩ and b=⟨−4,0,−5⟩?

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Answer 1 · 2017-05-25T18:13:55

$+-(-10/(7sqrt5),-9/(7sqrt5),8/(7sqrt5)).$

We know that, $vec x xx vec y$ is $bot$ to $vec x and vec y.$

Further, for $vec v!=vec 0, +-vec v/||vec v||$ are the unit vectors along $vec v.$

Now, $vec a xx vec b =|(i,j,k),(-1,2,1),(-4,0,-5)|$

$=(-10-0)i-(5-(-4))j+(0-(-8)k,$

$:. vec a xx vecb=-10i-9j+8k=(-10,-9,8).$

$:. ||vec a xx vecb||=sqrt{(-10)^2+(-9)^2+8^2}=sqrt245=7sqrt5.$

Therefore, $+-(-10/(7sqrt5),-9/(7sqrt5),8/(7sqrt5))$ are the desiredunit vectors.

Answer 2 · 2017-05-25T23:00:10

$hatc=+-sqrt5/35((-10),(-9),(8))$

We need to find a vector, $vecc$ , such that $veccbotveca$ and $veccbotvecb$ .

As such $vecc * veca=0$ and $vecc*vecb=0$

Let $vecc=((x),(y),(z))$

Then $-x+2y+z=0$ and $-4x-5z=0$

Let $z=1$

$-x+2y=-1$

$-4x=5$

$x=-5/4$

$-(-5/4)-2y=-1$

$2y=-9/4$

$y=-9/8$

$vecc=+-1/8((-10),(-9),(8))$

$c=sqrt(x^2+y^2+z^z)$

$c=sqrt((-5/4)^2+(-9/8)^2+1)=(7sqrt5)/8$

$hatc=1/c vecc$

$hatc=8/(7sqrt5)xx+-1/8((-10),(-9),(8))$

$hatc=+-sqrt5/35((-10),(-9),(8))$