# a=3i+4j and b=2i+j+3k |a| = ?  |b|= ? |a-b| = ? |axb| = ? Hello!

I think that i, j and k are mutually orthogonal unit vectors. In this case the length of a vector `x*i + y*j + z*k` is `sqrt(x^2+y^2+z^2).`

So

`|a| = sqrt(3^2+4^2+0^2) = sqrt(25) = 5,`

`|b| = sqrt(2^2+1^2+3^2) = sqrt(14) approx 3.74.`

Next, `a-b = (3i-2i) + (4j-j) +...

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Hello!

I think that i, j and k are mutually orthogonal unit vectors. In this case the length of a vector `x*i + y*j + z*k` is `sqrt(x^2+y^2+z^2).`

So

`|a| = sqrt(3^2+4^2+0^2) = sqrt(25) = 5,`

`|b| = sqrt(2^2+1^2+3^2) = sqrt(14) approx 3.74.`

Next, `a-b = (3i-2i) + (4j-j) + (0-3k) = i+3j-3k,`

`|a-b| = sqrt(1^2+3^2+3^2) = sqrt(19) approx 4.36.`

The last, axb. This is probably the vector product. Using formal determinant, it can be written as follows (`a_n` and `b_n` are the coefficients at i, j and k):

`| i j k |`
`| a_1 a_2 a_3 |`
`| b_1 b_2 b_3 |,`

which is equal to

`(a_2*b_3-b_2*a_3)*i + (a_1*b_3-b_1*a_3)*j + (a_1*b_2-b_1*a_2)*k.`

For these a and b

`axb = (4*3-1*0)*i + (3*3-2*0)*j+(3*1-2*4)*k = 12i+9j-5k`

and

`|axb| = sqrt(12^2+9^2+5^2) = sqrt(250) approx 15.81.`

Approved by eNotes Editorial Team