Vector Algebra

If four points and
are coplanar, then

Now,

$\left|\vec{a}+\vec{b}\right|

Let
Then,


Now,


Similarly,
and

Step 1: Recall the formula for the area of a parallelogram formed by two adjacent vectors.
The area of a parallelogram formed by two adjacent vectors and is given by the magnitude of their cross product: We know that .
Step 2: Calculate the magnitude of vector .
Given , the magnitude is:
Step 3: Use the relationship and the given answer to infer a different value for .
If the area is , then . Using the formula: This contradicts the given . Alternative Scenario (Assuming a typo in ): Let's assume, for the sake of matching the answer, that the vector or the dot product was different. If the area is , and , , then: If , for instance if with , this would lead to the correct area. However, this contradicts the given .

Step 1: Interpret the given conditions.
We have as unit vectors, so . The conditions are:
, so is perpendicular to .
, so is perpendicular to .
The angle between and is , so:
Step 2: Analyze the geometric implications.
Since is perpendicular to both and , must be perpendicular to the plane spanned by and . The cross product is a vector perpendicular to both and , so must be parallel to . Thus, we hypothesize: where is a scalar, and since is a unit vector, we need to determine .
Step 3: Compute the magnitude of .
The magnitude of the cross product is: If , then: Since : Thus:
Step 4: Verify the result.
Check orthogonality: and , so satisfies .
Check the unit vector condition: The magnitude of is 1, as computed above.
Option (D) matches exactly.

Step 1: Set up the condition for non-coplanarity.
Three vectors , , and are non-coplanar if they are linearly independent. Since are non-coplanar (hence linearly independent), we can express the vectors in the basis . Form the matrix with these vectors as rows: The vectors are non-coplanar if the determinant of this matrix is non-zero.
Step 2: Compute the determinant.
The matrix is upper triangular, so the determinant is the product of the diagonal elements: The vectors are linearly dependent (hence coplanar) when the determinant is zero:
Step 3: Analyze the result.
If , the vectors become , , . The second and third vectors are scalar multiples of each other ( and ), so they are coplanar.
If , the vectors become , , . The third vector is the zero vector, making them coplanar.
For all other , the determinant , so the vectors are linearly independent and non-coplanar.

Step 4: Determine the number of exceptional values.
There are two values of ( and ) where the vectors are coplanar. The vectors are non-coplanar for all except these two values.
Step 5: Select the correct answer.
The vectors are non-coplanar for all except and , which matches option (C) "all except two values of ".