Perpendicular Vectors?

Question

I asked this question yesterday but need a clearer answer. Please don't use the cross product we haven't learned it yet. Thanks

Find a vector C perpendicular to A=i+j+k and B=i-k

Answer 1

A = 1i + 1j + 1k

B = 1i + 0j -- 1k

C = xi + yj + zk

x, y, z are unknowns.

Calculating dot products:

A.C = 1x + 1y + 1z

B.C = 1x + 0y -- 1z

If C_|_A, and C_|_B, then A.C = 0, and B.C = 0, therefore

1x + 1y + 1z = 0

1x + 0y -- 1z = 0

We only know that C must be perpendicular to A and B, but nothing about its magnitude, its magnitude is indefinite. Give one component an arbitrary value, we will be able to find the other 2.

Put z = a

1x + 1y  = --a

1x + 0y =  +a

x = a, y = -2a, z = a

C = a(1i -- 2j + 1k)

Answer 2

The cross product is the way you would practically approach this problem, but since you said you didn't learn it, we'll do it the hard way.

Let vector C = ai + bj + ck

Now you know that if the vector C is perpendicular to these other vectors, the dot product between the two must be zero.

So (ai+bj+ck) dot (i + j + k) = 0

And

(ai+bj+ck) dot (i-k) = 0

So a + b + c  = 0 from the first one

And

a - c = 0 from the second one

So a = c, a + b + c = 0

2a + b = 0 by substitution

b = -2a

So there are an infinite number of vectors that satisfy this. There is one free variable and 2 dependent ones. So pick any constant for any of these variables and you will get a vector that satisfies the condition of orthogonality.

So let a = c = 1

Then b = -2(1) = -2

So one vector C might be <1,-2,1>, but there are an infinite number of multiples of this. If, perhaps, you wanted a unit vector that could be scaled, divide by the magnitude which is sqrt(1^2+(-2)^2+1^2) = sqrt(6)

<1/sqrt(6),-2/sqrt(6),1/sqrt(6)>