Take any vector A in (R^3. Let W be the set of all vectors B in (R^3 where B.A = 0. Discuss whether W is a subspace of (R^3 or not.
To determine whether W is a subspace of R^3, we need to ensure that it satisfies three conditions:
1) W contains the zero vector
2) W is closed under vector addition
3) W is closed under scalar multiplication.
1) W contains the zero vector:
Let A = (a1, a2, a3). To satisfy B.A = 0, we can choose B = (0, 0, 0), which is the zero vector. Therefore, W contains the zero vector.
2) W is closed under vector addition:
Consider two vectors B1 and B2 in W, such that B1.A = B2.A = 0.
Now we need to show that the sum of B1 and B2, denoted as B1 + B2, also satisfies B1 + B2 . A = 0.
(B1 + B2) . A = B1 . A + B2 . A (by the distributive property)
= 0 + 0 (since B1.A = B2.A = 0)
= 0
Therefore, B1 + B2 also satisfies B1 + B2 . A = 0, and W is closed under vector addition.
3) W is closed under scalar multiplication:
Consider a vector B in W and a scalar c. We need to show that cB also satisfies (cB).A = 0.
(cB).A = c(B.A) (by the distributive property)
= c(0) (since B.A = 0)
= 0
Therefore, cB also satisfies (cB).A = 0, and W is closed under scalar multiplication.
Since W satisfies all three conditions, we can conclude that W is a subspace of R^3.
To determine if the set W of all vectors B in R^3 where B·A = 0 is a subspace of R^3, we need to check three conditions:
1. The zero vector is in W:
Let's consider the zero vector, denoted as 0. Since 0·A = 0 for any vector A, the zero vector is included in W.
2. W is closed under vector addition:
Suppose we have two vectors B1 and B2 in W, such that B1·A = 0 and B2·A = 0. We need to show that the vector sum B1 + B2 is also in W. Let's calculate the dot product:
(B1 + B2)·A = B1·A + B2·A (using the distributive property of dot product)
= 0 + 0 (since B1·A = 0 and B2·A = 0)
= 0
So, the vector B1 + B2 satisfies the condition B·A = 0 and is also in W. Thus, W is closed under vector addition.
3. W is closed under scalar multiplication:
Consider any vector B in W, such that B·A = 0. We need to show that for any scalar c, the vector cB is also in W. Let's calculate the dot product:
(cB)·A = c(B·A) (using the distributive property of dot product)
= c(0) (since B·A = 0)
= 0
Hence, the vector cB satisfies the condition B·A = 0 and is in W. Therefore, W is closed under scalar multiplication.
Since W satisfies all three conditions, it can be concluded that the set W of all vectors B in R^3 satisfying B·A = 0 is a subspace of R^3.
To determine whether the set W of all vectors B in R^3 such that B.A = 0 is a subspace of R^3, we need to check if three conditions are satisfied: closure under addition, closure under scalar multiplication, and the presence of the zero vector.
1. Closure under addition: We need to show that if two vectors B1 and B2 are in W, then their sum B1 + B2 is also in W. Assume that B1.A = 0 and B2.A = 0. Now, we need to show that (B1 + B2).A = 0.
(B1 + B2).A = B1.A + B2.A [Using the distributive property of dot product]
= 0 + 0 [Since B1.A = 0 and B2.A = 0]
= 0
Thus, (B1 + B2).A = 0, which means that B1 + B2 is also in W. So, W satisfies closure under addition.
2. Closure under scalar multiplication: We need to show that if B is in W and c is any scalar, then cB is also in W. Assume that B.A = 0. Now, we need to show that (cB).A = 0.
(cB).A = c(B.A) [Using the scalar property of dot product]
= c(0) [Since B.A = 0]
= 0
Thus, (cB).A = 0, which means that cB is also in W. So, W satisfies closure under scalar multiplication.
3. Zero vector: The zero vector 0 is in W if it satisfies the condition 0.A = 0. Since the dot product of any vector with the zero vector is always 0, it follows that 0 is in W.
Since W satisfies all three conditions, we can conclude that it is indeed a subspace of R^3.