Question

Let \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \) be non-coplanar vectors. If \( \alpha \overrightarrow{d} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \), \( \beta \overrightarrow{a} = \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} \), then \( | \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} | = ? \)

• A. \( 1 \)
• B. \( 2 \)
• C. \( |\overrightarrow{a} - \overrightarrow{b} - \overrightarrow{c} | \)
• D. \( 0 \)

Hint:

When solving vector equations, check whether given vectors sum to zero. If the given system ensures that the vectors cancel out, then their resultant magnitude is zero.

Answer 1

The Correct Option is D

We are given the vector equations: \[ \alpha \overrightarrow{d} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \] \[ \beta \overrightarrow{a} = \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} \] Step 1: Express the sum of given vectors
Adding both equations: \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = (\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}) + (\overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d}) \] \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} \] \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = \overrightarrow{a} + \overrightarrow{d} + 2(\overrightarrow{b} + \overrightarrow{c}) \] Step 2: Finding the modulus
Since the given vectors are non-coplanar, their sum must satisfy: \[ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} = 0 \] Thus, the magnitude is: \[ | \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} | = 0 \]

Answer 2

We are given the vector equations: \[ \alpha \overrightarrow{d} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \] \[ \beta \overrightarrow{a} = \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} \] Step 1: Add the two given equations 

Adding both equations, we get: \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = (\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}) + (\overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d}) \] \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = \overrightarrow{a} + \overrightarrow{d} + 2(\overrightarrow{b} + \overrightarrow{c}) \] Step 2: Rearrange and isolate vectors 

Rearranging, we have: \[ (\alpha - 1) \overrightarrow{d} + (\beta - 1) \overrightarrow{a} = 2(\overrightarrow{b} + \overrightarrow{c}) \] Since \(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}, \overrightarrow{d}\) are non-coplanar vectors, this equation implies the right side must be expressible as a linear combination of \(\overrightarrow{a}\) and \(\overrightarrow{d}\). 
Step 3: Use non-coplanarity to find the sum 

For this to hold, the only possibility is: \[ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} = \vec{0} \] Therefore, the magnitude of the sum is: \[ |\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d}| = 0 \]