Hi,

How do you find a vector which is perpendicular to both a and b where:

a = 2i - 3j - 4k
b = 4i - 4j + k

Thank you.

Well, there are a variety of ways of doing it, depending on how much you know about vectors. Do you know about the dot product of two vectors, a .b =a_x b_x +a_y b_y +a_z b_z , which is 0 if a and b are perpendicular?

Use the vector (cross) product to get

a xb =-19i -18j +4k

Alternatively, consider the intersection of two planes

x .a =0; x .b =0

and solve for x , which will have a degree of freedom.

Kerwin

Hi,

Dan, I have heard of the dot product of two vectors. Kerwin, I have not learnt 'degree of freedom'.

Dan, but I don't know how to use this fact to find the answer to the question. Can you help?

In our case, a.b = (2x4) + (-3x-4) + (-4x1) = 8 + 12 - 4 = 16? Now what do we do?

Well, let x =xi +yj +zk , and substitute into

x.a =0; x.b =0

we have

2x-3y-4z=0
4x-4y+z=0

for x to be perpendicular to both a and b . Now we can fix one of x, y, z and solve the equation simultaneously to give the other 2 in terms of the one we fixed. This is known as a degree of freedom because you can choose the value of one variable freely. Now, say we fix x=1 and we find y=18/19, z=-4/19. (Obviously we want to avoid the case x =0 )

Kerwin

Thanks Kerwin, Dan for your explanations.

Here is what I think I have to do,

a = 2i - 3j - 4k
b = 4i - 3j + k

and let the perpendicular vector what we need to find the values of x,y,z, be c ,
c = xi + yj + zk

From what you said, that a vector is perpendicular to eachother, then,
b .c = a .c = 0 (because as you said Dan, cos90=0, thus at right angles to eachother)

a .c = (2x)i + (-3y)j + (-4z)k = 0
b .c = (4x)i + (-3y)j + (z)k = 0

Hence we can write as,
2x - 3y - 4z = 0
4x - 3y + z = 0

I lost you when you said, to let x=1 and then solve the 2 equations, why can you choose any value of x?

Well, first of all, a.c etc. are scalars, so there should not be any i, j, k in the answer.

Now, if you consider the two planes, they intersect at a line through origin. Any vector on the line will do the job. Another way of thinking about this is that we have 3 unknowns but only 2 equations, so we can only expressed 2 variables in terms of the third. Thus, you can take an arbitrary value of x and solve for y, z and still have a vector satisfying the required condition.

Kerwin

Thanks,
How do I prove the Cosine rule using Vector methods?

$O{A}^2=a.a$

$O{B}^2=b.b$

$A{B}^2=(a-b).(a-b)=a.a-2a.b+b.b$

which gives the required result when you substitute $a.b=ab\cos \theta$

Kerwin

Thanks Kerwin.

How do you prove that the angle in a semi-circle is a right angle by vector methods?

Let AC be the diameter, and O be the mid-point. Further, let B be a point on the circumference. Let OA, OB be a , b respectively and we are required to show that

(a -b ).(a +b )=0,

which is immediate because a² =b² =r² , where r is the radius of the circle.

Kerwin

Thanks Kerwin.

If A(2,1,7), B(-3,1,4), C(2,-1,5) how do you find the cosine of < ABC (angle ABC) using vector methods?

Do you find vector BC and BA? Then use $a.b=\left|a\right|\left|b\right|\cos \theta$?

Yes, that is the right method.

Thanks Kerwin.