# find a unit vector that is orthogonal to both i+j & i+k?

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find a unit vector that is orthogonal to both i+j & i+k?

I know that the length is sqrt(a^2 + a^2 + a^2) = asqrt(3) but I don’t know how to preceed to get the unit vector

Do this in two steps: first find a vector V that is orthogonal to both, then make it a unit vector by taking V / |V|.

Finding the orthogonal vector is simple: just use the cross-product, (i+j) x (i+k).

Once you calculate that, just take its length and divide by that, and you’ll naturally have a unit vector.

• Northstar

Find a unit vector that is orthogonal to both i+j and i+k.

Take the cross product. The resultant vector n will be orthogonal to both vectors.

n = <1, 1, 0> X <1, 0, 1> = <1, -1, -1>

Now we need to divide n by its magnitude so that we have a unit vector. Calculate the magnitude.

|| n || = √[1² + (-1)² + (-1)²] = √(1 + 1 + 1) = √3

Divide n by the magnitude to get the unit vector.

n / || n || = <1/√3, -1/√3, -1/√3>

• Anonymous

First, just find a vector that is orthogonal to the vectors (1,1,0) and (1,0,1). The dot product of two orthogonal vectors is 0.

Thus if we call our new vector (x,y,z), we have

(x,y,z) dot (1,1,0) = x + y = 0

(x,y,z) dot (1,0,1) = x + z = 0

Solving this yields

x = -y

x= -z

There are infinite solutions to this system of equations. Just pick x=1, and we get x=1, y=-1, z=-1.

Our vector is therefore (1,-1,-1). Just divide by the magnitude to get the unit vector. The magnitude is sqrt(3). So the unit vector is

(1/sqrt(3),-1/sqrt(3),-1/sqrt(3))

There ya go!

• joerling

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