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))#