if A=i+j, B=2i-3j+k and C=4j-3k
find
(i) (AxB)xC & Ax(BxC)
I'll start this out, using the determinant rule for vector cross products. You do the last part, and check my work for errors.
A x B =
|i j k|
|1 1 0|
|2 -3 1| = i -j -5k
(AxB) x C =
|i j k|
|1 -1 -5|
|0 4 -3| = 17i + 3j +4k
Well, it seems like you're attempting to put my mathematical skills to the test. Let's see if I can clown around with these calculations!
(i) First, let's find AxB:
To multiply two vectors, we need to find the cross product. So, AxB will be:
(AxB) = (i+j) x (2i-3j+k)
To calculate this, let's apply the cross product formula:
(AxB) = (i x (2i-3j+k)) + (j x (2i-3j+k))
Expanding this, we get:
(AxB) = (2i²-3ij+ki) + (2ji-3j²+jk)
Since i² = j² = k² = -1, we can simplify this further:
(AxB) = (2i-3ij+ki) + (2ji-3j+jk)
Combining like terms, we get:
(AxB) = 2i - 3ij + ki + 2ji - 3j + jk
Simplifying once again, we find:
(AxB) = (2i + ki + 2ji - 3j - 3ij + jk)
Now, let's multiply (AxB) by C:
((AxB)xC) = (2i + ki + 2ji - 3j - 3ij + jk) x (4j - 3k)
To calculate this, we apply the cross product once again:
((AxB)xC) = (2i x (4j - 3k)) + (ki x (4j - 3k)) + (2ji x (4j - 3k)) - (3j x (4j - 3k)) - (3ij x (4j - 3k)) + (jk x (4j - 3k))
Expanding this, we get:
((AxB)xC) = (8ij - 6ik) + (4kij - 3k²i) + (8j²i - 6ji²) - (12j² - 9jk) - (12ij² + 9ijk) + (4jk² - 3k³)
Since i² = j² = k² = -1, we can simplify this further:
((AxB)xC) = (8ij + 6ik) + (4kij + 3ki) + (8ji + 6j) - (12j - 9jk) - (12ij - 9jk) + (4jk - 3k)
Combining like terms, we have:
((AxB)xC) = (8ij + 4kij + 8ji - 12j - 12ij + 4jk) + (6ik + 3ki + 6j - 9jk - 9jk - 3k)
Simplifying even further, we find:
((AxB)xC) = (8ij - 8ij + 4kij - 12j - 12ij + 4jk) + (6ik + 3ki + 12j - 18jk - 3k)
Canceling out terms, we have:
((AxB)xC) = (4kij - 12ij + 4jk) + (6ik + 3ki - 18jk)
Finally, if you want to find Ax(BxC), let's retrace our steps and get there:
First, let's find BxC:
To calculate this, we use the cross product formula:
(BxC) = (2i-3j+k) x (4j-3k)
Expanding this, we get:
(BxC) = (2i x (4j-3k) - 3j x (4j-3k) + k x (4j-3k)
Simplifying, we find:
(BxC) = (8ij - 6ik) - (12j² - 9jk) + (4jk - 3k²)
Since we know that i² = j² = k² = -1, we can simplify this further:
(BxC) = (8ij + 6ik) - (12j - 9jk) + (4jk + 3k)
Combining like terms, we have:
(BxC) = (8ij + 6ik - 12j + 9jk + 4jk + 3k)
Simplifying once again, we find:
(BxC) = (8ij + 9jk + 4jk + 6ik - 12j + 3k)
Now, to find Ax(BxC), we use a similar process as before:
(Ax(BxC)) = (i+j) x (8ij + 9jk + 4jk + 6ik - 12j + 3k)
Expanding this, we find:
(Ax(BxC)) = (i x (8ij + 9jk + 4jk + 6ik - 12j + 3k)) + (j x (8ij + 9jk + 4jk + 6ik - 12j + 3k))
Simplifying, we have:
(Ax(BxC)) = (8i²j + 9ijk + 4ijk + 6i²k - 12ij² + 3ikj) + (8ij² + 9j²k + 4j²k + 6ij²k - 12j³ + 3jk²)
Since i² = j² = k² = -1, we can simplify even further:
(Ax(BxC)) = (-8ij + 9ijk + 4ijk - 6ik - 12j - 3ikj) + (8ij - 9jk - 4jk + 6j - 12j³ - 3k)
Canceling out terms, we find:
(Ax(BxC)) = (-12j - 12j³ + 6j) + (-3k)
Simplifying, we have:
(Ax(BxC)) = (-12j - 12j³ + 6j - 3k)
So, putting it all together:
(i) (AxB)xC = (4kij - 12ij + 4jk) + (6ik + 3ki - 18jk)
Ax(BxC) = (-12j - 12j³ + 6j - 3k)
I hope you enjoyed my circus act of calculations!
To find (AxB)xC, we'll first find the cross product of A and B, and then take the cross product of the result with C.
(i) (AxB)xC:
Step 1: Find the cross product of A and B.
AxB = (i + j) x (2i - 3j + k)
To find the cross product, we can use the following formula:
AxB = (AyBz - AzBy) i + (AzBx - AxBz) j + (AxBy - AyBx) k
By substituting the values:
AxB = (1*(-3) - 0) i + (0 - 2*1) j + (1*3 - 1*0) k
= -3i - 2j + 3k
Step 2: Take the cross product of AxB with C.
(AxB)xC = (-3i - 2j + 3k) x (4j - 3k)
Using the same formula:
(AxB)xC = ((-3)*(0) - (3)*(-3)) i + ((3)*(4) - (-3)*(0)) j + ((-3)*(-3) - (-2)*(4)) k
= 9i + 12j + 7k
Therefore, (AxB)xC = 9i + 12j + 7k.
(ii) Ax(BxC):
Step 1: Take the cross product of B and C.
BxC = (2i - 3j + k) x (4j - 3k)
Using the cross product formula:
BxC = ((-3)*(0) - (1)*(4)) i + ((2)*(4) - (0)*(-3)) j + ((2)*(-3) - (-3)*(0)) k
= -4i + 8j - 6k
Step 2: Find the cross product of A with BxC.
Ax(BxC) = (i + j) x (-4i + 8j - 6k)
Using the cross product formula:
Ax(BxC) = ((1)*(8) - (0)*(0)) i + ((0)*(-4) - (1)*(-6)) j + ((1)*(-4) - (1)*(8)) k
= 8i + 6j - 12k
Therefore, Ax(BxC) = 8i + 6j - 12k.
In summary,
(i) (AxB)xC = 9i + 12j + 7k,
(ii) Ax(BxC) = 8i + 6j - 12k.
To find (AxB)xC, we need to find the cross product of A and B first, and then take the result of that cross product and calculate its cross product with vector C.
Step 1: Calculate the cross product of vectors A and B.
The cross product of two vectors (a1, a2, a3) and (b1, b2, b3) is given by the following formula:
(a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)
In our case, A = (i + j) and B = (2i - 3j + k).
Using the formula, we can calculate the cross product of A and B:
A x B = ((1)(1) - (1)(-3), (1)(2) - (1)(1), (1)(-3) - (1)(2))
= (1 + 3, 2 - 1, -3 - 2)
= (4, 1, -5)
Step 2: Calculate the cross product of the result from Step 1 and vector C.
Using the same formula, we can calculate the cross product of the result from Step 1 and C:
(A x B) x C = ((4)(0) - (-5)(3), (-5)(4) - (1)(0), (1)(3) - (4)(4))
= (15, -20, -13)
Therefore, (AxB)xC = (15, -20, -13).
Now, let's find Ax(BxC).
Step 3: Calculate the cross product of vectors B and C.
Using the formula mentioned earlier, we can calculate the cross product of B and C:
B x C = ((-3)(4) - (0)(-3), (0)(4) - (2)(-3), (2)(3) - (-3)(-3))
= (-12, 6, 15)
Step 4: Calculate the cross product of A and the result from Step 3.
Using the formula again, we can calculate the cross product of A and the result from Step 3:
A x (B x C) = ((1)(15) - (1)(6), (1)(-12) - (1)(15), (1)(6) - (1)(-12))
= (9, -27, 18)
Therefore, Ax(BxC) = (9, -27, 18).
In summary:
(AxB)xC = (15, -20, -13)
Ax(BxC) = (9, -27, 18)