Vectors

Step 1: Compute Step 2: Cross product Use determinant form: Compare with:

Condition: The equality holds if and only if vectors and are collinear and in the same direction — that is, for some scalar , or for , or one of the vectors is a zero vector.

Given:

Since , for collinearity in the same direction we assume: Substituting:

Comparing components:

• :
• : — consistent
• :

So, the value of is:

Verification:

If , then

Now compute magnitudes:

Hence, verified. The correct value is:

Concept: If vector is perpendicular to both and , then it must be parallel to the cross product:

Given:

Compute :

Simplify:

All components are divisible by 17: So,

Given: . Calculate magnitude of :

So,

Condition: makes an obtuse angle with

Thus, the correct choice is:

Check if this matches any option:

Option (d):

So, it is the same vector but scaled by 2:

Check orthogonality:

So, is indeed perpendicular to both and , and its y-component is negative → obtuse with .

Conclusion:

The correct direction and properties match option (d), even though the magnitude is doubled. This implies either:

• The magnitude in the question should have been
• Or the options should have been scaled appropriately to magnitude 14

✅ Final Answer: (Assuming either a typo in question's magnitude or uniform scaling in options.)

We are given the vector expression:
Step 1: Understanding the vector sum
This expression represents a weighted sum of the vectors and , where the weights are the magnitudes of the other vector. This type of vector sum is known to produce a vector that is in the direction of the angle bisector of the two vectors.
Step 2: Geometric Interpretation
- The vector lies in the plane formed by and . - The weights assigned to and ensure that the resultant vector is directed along the angle bisector of the angle between and .
Step 3: Conclusion
Since the given expression aligns with the well-known angle bisector theorem in vector form, the vector is parallel to the bisector of the angle between and . Thus, the correct answer is:

Step 1: Finding the Equation of the Line Passing Through and The parametric equation of the line passing through points and is given by: Expanding the terms:
Step 2: Finding the Equation of the Plane Passing Through and The normal vector to the plane is found by taking the cross-product of vectors and : Taking the cross-product: Thus, the equation of the plane is:
Step 3: Finding the Intersection of the Line and the Plane Substituting , into the plane equation: Substituting into the parametric equations: Thus, the position vector of is:

Step 1: Use Scalar Projection Formula
The magnitude of the projection of on is:

Step 2: Compute Dot Product
Let . Then:

Step 3: Compute Magnitude of

Step 4: Final Answer

Step 1: Define the position vectors.
Let: $ is: \) is: \) onto .
The projection of onto is given by: \) : \) : \) to is: \) .
Step 4: Compute .
Here, , , and . Therefore: \) $