Vector Algebra

Given that and are unit vectors: Now, So, Therefore, So,

The projection of on is: Here, Dot product: So,

Let’s use position vectors: So,

We are given: \begin{itemize} \item are unit vectors. \item \item \item Angle between and is \end{itemize} Step 1: Geometrical Meaning Since is perpendicular to both and , it must be perpendicular to the plane containing and . This implies: Step 2: Use Magnitudes Take magnitude on both sides: Given is a unit vector: Since and are unit vectors and , we have: Step 3: Final Answer Hence Proved. \quad

We are given: - , so (since magnitude is non-negative). - . - . We need to find the magnitude of . The magnitude of the sum of two vectors is given by: Expanding the dot product: Substituting the given values: - , - , - . So: Taking the square root (since magnitude is non-negative):

We are given that , and we need to find the angle between the vectors and . - The magnitude of the cross product is given by: - The dot product is given by: According to the problem: Assuming and are non-zero (as vectors have magnitude), we can divide both sides by (provided ):
Divide both sides by (assuming , i.e., ):
The angle for which is (in the first quadrant, where is between and ).
Verification
- If , then .
- .
- .
- Thus, , which satisfies the given condition.
Other angles like (where and ) would make , but , which does not hold unless or is zero, contradicting the problem's intent. Hence, is correct.