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

  • Fornoob Team’s answer

    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

    bypass Product: -8i-4j+4k reduces nicely to -2i-j+ok. additionally, opposite the signs and warning signs to get one pointing interior the alternative direction (2i+j-ok). Divide those by way of radical(6) to get unit vector lengths (2/rad(6)i+rad(6)j-rad(6)ok…. Double verify that by way of checking dot product against the two vectors. Dot products between the answer unit vectors and the two vectors interior the equation might desire to be 0. additionally, do no longer ignore your hats…. substantial: there is in elementary terms one unit vector in each direction. in simple terms multiplying the vector by way of 2 is erroneous (it is now no longer a unit vector) and in case you diverse the vector by way of a scalar (case in point 2 and then dividing by way of it incredibly is new magnitude in elementary terms gets you a similar vector you had. as an occasion, i+j+ok/rad(3) and 2i+2j+2k/rad(12) ARE a similar VECTOR!!! If i+j+ok/rad(3) became an orthogonal vector, the only different othogonal unit vector is interior the alternative direction or -i-j-ok/rad(3).

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