- #1

ND3G

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**Use two methods to determine a unit vector perpendicular to both (2,1,-3) and (-1, 7, 4)**

**Using Cross Product:**

Let v be a perpendicular vector to the given vectors

v = a*b

= (2,1,-3)(-1,7,4)

=((1)(4)-(7)(-3), (-3)(-1)-(4)(2), (2)(7)-(-1)(1))

=(4+21, 3-8, 14+1)

=(25, -5, 15)

= (5, -1, 3)

= |(5,-1, 3)|

= sqr(25+1+9

=sqr(35)

Let u be a unit vector in the direction of (5, -1, 3)

u = v/|v| = 1/sqr(35)(5, -1, 3) = (5/sqr(35), -1/sqr(35), 3/sqr(35))

**Using dot product:**

Let p = (x, y, z) represent the vector that is perpendicular to the vectors.

a = (2, 1, -3) and b = (-1, 7, 4)

Then p*a = 0 and p*b = 0

(x, y, z)(2, 1, -3) = 0 and (x, y, z)(-1, 7, 4) = 0

2x + y - 3z = 0

-x + 7y + 4Z = 0

Isolate x in equation (1) to get x = -y/2 + 3z/2

Substitute this into equation (2)

-(-y/2 + 3z/2) + 7y +4z = 0

y/2 - 3z/2 +7y +4z = 0

z = 0 --> substitute into equation (1) gives 2x + y = 0 and so x = -y/2

If you let x = 1, then y = -2

Therefore p = (x, y, z) = (1, -2, 0), which gives a different vector from the one found using cross product.

Determine the magnitude of this vector and then find a unit vector as required.

|(1, -2, 0)| = sqr(1+4) = sqr(5)

Let u be a unit vector in the direction of (1, -2, 0)

u = v/|v| = 1/sqr(5)(1, -2, 0) = (1/sqr(5), -2/sqr(5), 0)

**The two answers should match up but they don't and I am not sure what I am doing wrong. I have been over it a number of times but I just can not see where the error lies. Any help would be greatly appreciated**