Question

If \(\overset{̄}{a}\) and \(\overset{̄}{b}\) are unit vectors perpendicular to each other, then \([\overset{̄}{a}+(\overset{̄}{a}\times \overset{̄}{b})\,\overset{̄}{b}+(\overset{̄}{a}\times \overset{̄}{b})\,(\overset{̄}{a}\times \overset{̄}{b})] = \cdots\)

• A. \(-1\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

Subtract the third vector from the first two rows; the triple product is unchanged.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Let \(\bar c = \bar a \times \bar b\). The scalar triple product \([\bar a + \bar c,\ \bar b + \bar c,\ \bar c]\) is a determinant, and it does not change when we subtract a multiple of one row from another.

Step 2: Simplify:
Subtract the third row \(\bar c\) from each of the first two rows:
\[ [\bar a + \bar c,\ \bar b + \bar c,\ \bar c] = [\bar a,\ \bar b,\ \bar c] \]

Step 3: Evaluate:
\[ [\bar a\ \bar b\ \bar c] = \bar a\cdot(\bar b\times\bar c) = (\bar a\times\bar b)\cdot\bar c = |\bar a\times\bar b|^2 \]
For perpendicular unit vectors, \(|\bar a\times\bar b| = |\bar a||\bar b|\sin 90^{\circ} = 1\), so the value is 1.

Step 4: Why the other options are wrong.
The value -1 would need a negative square. Values 2 and 3 come from adding extra terms, such as \(\bar c\cdot\bar c\) more than once, which cancel out in the determinant.

Final Answer:
The value is 1, option (B). \[ \boxed{1} \]