if A= i+j, B= 2i-3j+k and C= 4j-3k
find: Ax(BxC)
pls help with solution.
A=[1,1,0]
B=[2,-3,1]
C=[4,0,-3]
Let's do BxC
I use a very simple algorith to form the cross-product of two vectors.
- write the two vectors above each other
2 -3 1
4 0 -3
- for the first number, with your pinkie or with a pencil, block off the first column and find the right cross-product of the remaining square matrix, that is, (-3)(-3) - (0)(1) = 9
-for the second number, with your pinkie or withe a pencil, block off the middle column and find the negative right cross-product of the remaining matrix, that is
-( (2)(-3) - (1)(4) ) = 10
- and finally for the third number, block off the third column and find the right cross-product of the remaining square matrix, that is, (2)(0) - (-3)(4) = 12
So BxC = [9,10,12]
( I always check by taking the dot product of this with the two original vectors, you should get zero)
Now repeat by taking [1,1,0]x[9,10,12]
I got [12, -12, 1} or 12i - 12j + k
To find the cross product Ax(BxC), we need to first find the cross product of B and C, and then take the dot product of A with the result.
Step 1: Find the cross product of B and C.
To find the cross product of two vectors, we can use the determinant method.
Cross product of B and C:
BxC = (2i - 3j + k) x (4j - 3k)
Using the determinant method, we can expand the cross product as follows:
BxC = [(2)(-3) - (-3)(4)]i + [(2)(4) - (k)(-3)]j + [(2)(-3) - (4)(-3)]k
Simplifying further:
BxC = (-6 - (-12))i + (8 + 3k)j + (-6 - (-12))k
= (6)i + (8 + 3k)j + (6)k
= 6i + (8 + 3k)j + 6k
Step 2: Take the dot product of A with BxC.
To take the dot product, we need to multiply the corresponding components of the two vectors and add them together.
Ax(BxC) = (i + j) • (6i + (8 + 3k)j + 6k)
Multiplying the corresponding components:
Ax(BxC) = (1)(6) + (1)(8 + 3k) + (0)(6)
= 6 + 8 + 3k
= 14 + 3k
Therefore, Ax(BxC) = 14 + 3k.
To find Ax(BxC), we need to properly distribute the multiplication and perform the vector operations.
First, let's find BxC.
BxC can be obtained by taking the cross product of vectors B and C. The cross product of two vectors, denoted by the "x" symbol, is given by the following formula:
B x C = (B2C3 - B3C2)i - (B1C3 - B3C1)j + (B1C2 - B2C1)k
Now, substituting the values of B and C into the formula:
B x C = ((2)(4) - (-3)(-3))i - ((2)(-3) - (-3)(0))j + ((2)(0) - (4)(-3))k
Simplifying the calculations:
B x C = (8 - 9)i - (-6 - 0)j + (0 + 12)k
= -1i - (-6)j + 12k
= -i + 6j + 12k
Now that we have BxC, we can find Ax(BxC).
Ax(BxC) is obtained by taking the dot product of vector A and the vector BxC. The dot product, denoted by the "." symbol, is given by the following formula:
Ax(BxC) = Ai(BxC)i + Aj(BxC)j + Ak(BxC)k
Now, substituting the values of A and BxC into the formula:
Ax(BxC) = (i + j)(-i + 6j + 12k)i + (i + j)(-i + 6j + 12k)j + (i + j)(-i + 6j + 12k)k
Expanding and simplifying the calculations:
Ax(BxC) = -i^2 + 6ij + 12ik - i^2 + ij + 6j^2 + 12jk + i^2 - 6ij - 12ik + ij - 6j^2 - 12jk
Note that i^2 = -1, j^2 = -1, and k^2 = -1:
Ax(BxC) = -(-1) - 6ij + 12ik - (-1) + ij - 6 - 12jk + (-1) + 6ij + 12ik + ij - 6 - 12jk
Simplifying the calculations further:
Ax(BxC) = 1 + 6ij + 12ik + 1 + ij - 6 - 12jk - 1 - 6ij - 12ik + ij - 6 - 12jk
= 2ij + 2ij - 12jk - 12jk
Combining like terms:
Ax(BxC) = 4ij - 24jk
Therefore, Ax(BxC) = 4ij - 24jk.