For the following three vectors, what is 3C·(2A multiplied by Bvec)?
A = 2.00ihat + 3.00jhat - 6.00khat
Bvec = -3.00ihat + 3.00jhat + 2.00khat
C = 9.00ihat - 8.00jhat
-54.00
To find the result of 3C · (2A multiplied by Bvec), we need to perform vector operations, including scalar multiplication, dot product, and vector addition.
Let's break it down step by step:
1. Scalar multiplication of A and Bvec:
Multiply each component of A by 2 and each component of Bvec by 2:
2A = 2(2.00ihat) + 2(3.00jhat) + 2(-6.00khat)
= 4.00ihat + 6.00jhat - 12.00khat
We now have the vector 2A.
2. Dot product of 3C and 2A:
Multiply each corresponding component of 3C and 2A and sum them up:
3C · 2A = (9.00ihat - 8.00jhat) · (4.00ihat + 6.00jhat - 12.00khat)
Multiply the corresponding components:
3C · 2A = (9.00 * 4.00) + (-8.00 * 6.00) + (0 * -12.00)
= 36.00 + (-48.00) + 0
= -12.00
The dot product of 3C and 2A is -12.00.
3. Multiply the result by Bvec:
Multiply the dot product result, -12.00, by each corresponding component of Bvec:
3C · (2A multiplied by Bvec) = (-12.00)(-3.00ihat + 3.00jhat + 2.00khat)
Multiply the dot product result by each component:
3C · (2A multiplied by Bvec) = (-12.00 * -3.00)ihat + (-12.00 * 3.00)jhat + (-12.00 * 2.00)khat
= 36.00ihat + -36.00jhat + -24.00khat
Simplifying the result, we get:
3C · (2A multiplied by Bvec) = 36.00ihat - 36.00jhat - 24.00khat
Therefore, the result of 3C · (2A multiplied by Bvec) is 36.00ihat - 36.00jhat - 24.00khat.
To find the result of the expression 3C·(2A multiplied by Bvec), we need to calculate 2A multiplied by Bvec first, and then find the dot product with C.
Let's calculate 2A multiplied by Bvec:
2A = (2.00)(2.00ihat + 3.00jhat - 6.00khat)
= 4.00ihat + 6.00jhat - 12.00khat
Now, multiply 2A by Bvec:
2A multiplied by Bvec = (4.00ihat + 6.00jhat - 12.00khat) multiplied by (-3.00ihat + 3.00jhat + 2.00khat)
To multiply these vectors, we will use the distributive property of multiplication:
2A multiplied by Bvec = (4.00ihat + 6.00jhat - 12.00khat)(-3.00ihat + 3.00jhat + 2.00khat)
= (-12.00i - 18.00j + 36.00k)(-3.00i + 3.00j + 2.00k)
= (-12.00)(-3.00)i + (-12.00)(3.00)j + (-12.00)(2.00)k
+ (-18.00)(-3.00)i + (-18.00)(3.00)j + (-18.00)(2.00)k
+ (36.00)(-3.00)i + (36.00)(3.00)j + (36.00)(2.00)k
= 36.00i + 54.00j - 108.00k + 54.00i - 54.00j - 36.00k - 108.00i
+ 108.00j + 72.00k
= (36.00i + 54.00i - 108.00i) + (54.00j - 54.00j + 108.00j) + (-108.00k - 36.00k + 72.00k)
= -18.00i + 108.00j - 72.00k
Now, let's find the dot product of this result with C:
3C·(2A multiplied by Bvec) = (3.00)(-18.00i + 108.00j - 72.00k)·(9.00i - 8.00j)
= (3.00)(-18.00)(9.00) + (3.00)(108.00)(-8.00) + (3.00)(-72.00)(0)
- (3.00)(-18.00)(0) - (3.00)(108.00)(8.00) - (3.00)(-72.00)(-8.00)
= -486.00 - 2592.00 + 0 + 0 + 2592.00 + 1728.00
= -486.00 + 2592.00 + 1728.00
= 2834.00
Therefore, 3C·(2A multiplied by Bvec) is equal to 2834.00.