please help in math?

Question

What is the easiest way to solve this vector product:
(5i-j+3k) x (i+3j+2k).
Do u know any easiest way or tricks to solve this problem?
please tell me.

this is a cross product problem

Answer 1

The easiest way is with technology: graphing calculators or computer algebra systems. With Maple I find:

> with(linalg):
Warning, new definition for norm
Warning, new definition for trace
> crossprod([5,-1,3],[1,3,2]);
[-11, -7, 16]

"By hand" methods are (1) memorize the key formula

[a,b,c]x[x,y,z]=[b*z-c*y, c*x-a*z, a*y-b*x]

or (2) if you know determinants, you find the determinant of the matrix with i,j,k in the first row, one vector in the 2nd row and the other vector in the third row.

Answer 2

NOTE: I think guys above want to give you the answer for a scalar product. Your problem is the cross product (vectorial product)


If you know how to find the determinant of a matrix you can make this array:
i j k
5 -1 3
1 3 2

(first row is i, j,k, second is coefficients of vector 5i-j+3k and third is the coefficients of i+3j+2k)

Now you find the determinant:
i(-1*2-3*3)-j(5*2-1*3)+k(5*3-1*-1) = -11i -7j +16k

Or if you want a general formula: for (ai+bj+ck)x(di+ej+fk)
determinant of:
i j k
a b c
d e f
makes the product:

(ai+bj+ck)x(di+ej+fk)=
(bf-ce)i -(af-cd)j + (ae-bd)k

Answer 3

yep you just use the vector dot product
<5x 1>, <-1x 3>, <3x2>

the resultant vector will be <5, -3, 6>
or writing it out 5i-3j+6k

you just multiply the like vectors (i's times the i's yada yada)

Answer 4

I think its everything times everything.
5i^2+15ij+10ik-ij-3j^2-2jk+3ik+9jk+6k^2
now combine like terms
5i^2 +14ij +13ik -3j^2 +7jk +6k^2

Answer 5

I HAVE ALWAYS FOUND IT VERY USEFUL TO USE AN 'IJK TRIANGLE' SUCH THAT:

.....I

K........J


MOVING CLOCKWISE AROUND THE TRIANGLE IS POSITIVE, AND CCW IS NEGATIVE.

FOR INSTANCE: i x j = +k but j x i = -k
LIKEWISE: k x i = +j and i x k = -j

THEREFORE, YOUR ANSWER WOULD BE, TAKING TERM BY TERM:

0 +15j -10j +k + 0 -6i +3j -9i +0 = -15i +8j +k