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We are given a vector \(\langle 1, a, b \rangle\) that needs to be orthogonal to the vector \(\langle 6, -2, -6 \rangle\). 

Two vectors are orthogonal if their dot product is zero. The dot product of \(\langle 1, a, b \rangle\) and \(\langle 6, -2, -6 \rangle\) is calculated as follows:

\[
1 \cdot 6 + a \cdot (-2) + b \cdot (-6) = 0
\]

This simplifies to:

\[
6 - 2a - 6b = 0
\]

Now, solve for \(b\) in terms of \(a\):

\[
6 - 2a - 6b = 0 \quad \Rightarrow \quad -2a - 6b = -6 \quad \Rightarrow \quad 2a + 6b = 6
\]

\[
a + 3b = 3
\]

Solving for \(b\):

\[
b = \frac{3 - a}{3}
\]

Therefore, the expression for \(b\) is \(\frac{3 - a}{3}\), meaning the correct choice is **B**, where \(b = \frac{3 - a}{3}\) and \(a\) is any real number.

Would you like more details or have any further questions?

Here are 5 related questions to expand this topic:

1. What are the properties of orthogonal vectors in 3D space?
2. How do you generalize the dot product formula for vectors in \(n\)-dimensional space?
3. What is the geometric interpretation of the dot product being zero?
4. How would you calculate the orthogonal projection of one vector onto another?
5. What are the applications of orthogonality in real-world problems?

**Tip**: To check if two vectors are orthogonal, always calculate their dot product—it must be zero.

Let a and b be real numbers. Find all vectors ⟨1, a, b⟩ orthogonal to ⟨6, −2, −6⟩. What are all the vectors that are orthogonal to ⟨6, −2, −6⟩?

Learn how to find all vectors ⟨1, a, b⟩ orthogonal to the vector ⟨6, −2, −6⟩. This problem involves the dot product and orthogonality in 3D space, and is suitable for students in Grades 11-12.

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